Raju

Students often turn to ISC Mathematics Class 11 OP Malhotra Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b) to clarify doubts and improve problem-solving skills.

S Chand Class 11 ICSE Maths Solutions Chapter 26 Points and their Coordinates in 3-Dimensions Ex 26(b)

Question 1.
Find the coordinates of the points which divide the join of the points (2, -1, 3) and (4, 3, 1) in the ratio 3 : 4 internally.
Solution:
Let R be the point which divides the join of points A(2, -1, 3) and B(4, 3, 1) in the ratio 3 : 4 internally.

Question 2.
Find the coordinates of the points which divide the line joining the points (2, -4, 3), (-4, 5, -6) in the ratio
(i) 1 : -4
(ii) 2 : 1
Solution:
(i) Let the point P divides the line segment AB in ratio 1 :-4
Then coordinates of P are

Question 3.
Find the ratio in which the line joining the points (2, 4, 5),(3, 5, -4) is divided by the yz-plane.
Solution:
Let the point P divides the line segment joining A(2, 4, 5) and B(3, 5, -4) in the ratio k : 1

Since the line joining AB is divided by yz – plane
i.e. x-coordinates of point P is 0 .
\(\frac{3 k+2}{k+1}\) = 0
⇒ k = –\(\frac{2}{3}\)
Thus required ratio be k : 1 i.e. –\(\frac{2}{3}\) : 1 i.e. – 2 : 3.

Question 4.
The three points A(0, 0, 0), B(2, -3, 3), C(-2, 3, -3) are collinear. Find in what ratio each point divides the segment joining the other two.
Solution:
Let the point B (2, -3, 3) divides the line segment AC in the ratio k : 1 internally.
Then by section formula, we have
The coordinates of B are

∴ \( \frac{-2 k}{k+1}\) = 2
⇒ – 2k = 2k + 2
⇒ 4k = – 2
⇒ k = \(\frac{-1}{2}\)
and \(\frac{3 k}{k+1}\) = – 3
⇒ 3k = -3 k – 3
⇒ 6 k = – 3
⇒ k =-\(\frac{1}{2}\)
and \(\frac{-3 k}{k+1}\) = 3
⇒ – 3k = 3k + 3
⇒ k = –\(\frac{1}{2}\)
Thus the required ratio be k : 1 i.e. – 1 : 2.
Let the point C(-2, 3, -3) divides AB in the ratio λ : 1.
Then coordinates of C are

\(\frac{2 \lambda}{\lambda+1}\) = – 2 ⇒ 2λ = – 2λ – 2 ⇒ λ = –\(\frac { 1 }{ 2 }\)
\(\frac{-3 \lambda}{\lambda+1}\) = 3 ⇒ -3λ = 3λ + 3 ⇒ λ = –\(\frac { 1 }{ 2 }\)
and \(\frac{3 \lambda}{\lambda+1}\) = – 3 ⇒ 6λ = – 3
Thus required ratio be λ : 1 i.e. -1 : 2. Let the point A(0, 0, 0) divides line segment BC in the ratio p : 1.

Question 5.
Find the coordinates of the points which trisect AB given that A(2, 1, -3) and B (5, -8, 3).
Solution:
Let P and Q be the point of trisection of line segment AB.

Thus point P divides the line segment AB in the ratio 1 : 2.
Then coordinates of P are
\(\left(\frac{5+4}{1+2}, \frac{-8+2}{1+2}, \frac{3-6}{1+2}\right)\) i.e. P(3, -2, -1)
Also, point Q divides the line segment AB in the ratio 2 : 1.
Then coordinates of $\mathrm{Q}$ are
\(\left(\frac{10+2}{2+1}, \frac{-16+1}{2+1}, \frac{6-3}{2+1}\right)\) i.e. Q(4, -5, 1).

Question 6.
Find the coordinates of the point which is three-fifths of the way from (3, 4, 5) to (-2, -1, 0).
Solution:
Let the given points are A(3, 4, 5) and E(-2, -1, 0) and let point P is at \(\frac{3}{5}\)th of the way from A.
∴ P is at a \(\frac{2}{5}\)th of the way from B i.e. p divides line segment AB in the ratio 3 : 2.
Then by section formula, we have

Thus, the coordinates of point P are (0, 1, 2).

Question 7.
Show that the point (1, -1, 2) is common to the lines which join (6, -7), 0) to (16, -19, -4) and (0, 3, -6) to (2, -5, 10).
Solution:
Any point on line segment AB be

If AB and CD have a common point. Then P and Q coincide for some values of k and k’.

Thus eqns. (1), (2) and (3) are satisfied or consistent for k = \(\frac{-1}{3}\) and k’ = 1
putting k = \(\frac{-1}{3}\) in coordinates of P
we get, the required point be (1, -1, 2). Hence, the point P(1, -1, 2) is common to lines which join A(6, – 7, 0)
and B(16, -19, -4) and C(0, 3, -6) and D(2, -5, 10).

Question 8.
Find the lengths of the medians of the triangle whose vertices are A(2, -3, 1), B (-6, 5, 3), C (8, 7, – 7).
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of △ABC.

Question 9.
Find the point of intersection of the medians of the triangle with vertices (-1, -3, -4), (4, -2, -7), (2, 3, -8).
Solution:
Let the vertices of △ABC are A(-1, -3, -4); B(4, -2, -7) and C(2, 3, -8)
We know that the point of intersection of all medians of a triangle is called centroid of triangle.

Question 10.
Find the ratio in which the join of A(2, 1, 5) and B(3, 4, 3) is divided by the plane 2x + 2y – 2z = 1. Also, find the coordinates of the point of division.
Solution:
Let the point R divides the line segment PQ in the ratio k : 1 internally.

Then by section formula, we have coordinates of R are
\(\left(\frac{3 k+2}{k+1}, \frac{4 k+1}{k+1}, \frac{3 k+5}{k+1}\right)\)
Clearly it is given that, line segment PQ is divided by the plane
2x + 2y – 2z = 1 …(1)
Thus the point R lies on eqn. (1); we have
\(2\left(\frac{3 k+2}{k+1}\right)+2\left(\frac{4 k+1}{k+1}\right)-2\left(\frac{3 k+5}{k+1}\right)\) = 1
⇒ 6k + 4 + 8k + 2 – 6k – 10 = k + 1
⇒ 7k = 5 ⇒ k = \(\frac{5}{7}\)
Thus the required ratio be k : 1
i.e. \(\frac{5}{7}\) : 1 i.e. 5 : 7
Thus, required point of division be

Question 11.
The mid-points of the sides of astriangle are (1, 5, -1),(0, 4, -2) and (2, 3, 4). Find its vertices.
Solution:
Let the vertices of △ABC are A (x₁, y₁, z₁);
B (x₂, y₂, z₂) and
C (x₃, y₃, z₃).
It is given that D (1, 5, -1) ; E(0, 4, – 2) and F(2, 3, 4) are the mid-points of sides BC, CA and AB of △ABC.
Now D be the mid-point of BC.

On adding eqn. (1), (4) and (7); we have
x₁ + x₂ + x₃ = 3 …(10)
From eqn. (1) and eqn. (10); x₁ = 1
From (4) and (10); x₂ = 3
and From eqn. (7) and (10); x₃ = -1
On adding eqn. (2), (5) and (8); we have
y₁ + y₂ + y₃ = 12 …(11)
From (2) and (11); y₁ = 2
From (5) and (11); y₂ = 4
From (8) and (11); y₃ = 6
On adding eqn. (3), (6) and (9); we have
z₁ + z₂ + z₃ = 1 …(12)
From (3) and (12); z₁ = 3
From (6) and (12); z₂ = 5
From (9) and (12); z₃ = -7
Thus the required vertices of △ABC are (1, 2, 3) ;(3, 4, 5) and (-1, 6, -7).

Question 12.
Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex D.
Solution:
Given vertices of parallelogram ABCD are A(3, -1, 2); B(1, 2, -4); C(-1, 1, 2) and let the coordinates of fourth vertex D are (α, β ,γ).
Mid-point of AC = \(\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)\)
and Mid-point of BD = \(\left(\frac{1+\alpha}{2}, \frac{2+\beta}{2}, \frac{-4+\gamma}{2}\right)\)
Since ABCD be a parallelogram.
∴ diagonals of || gm ABCD bisect each other.
Thus, mid-point of AC = mid-point of BD

Thus the coordinates of fourth vertex D are $(1,-2,8)$.

Question 13.
What is the locus of a point for which
(i) x = 0
(ii) y = 0
(iii) z = 0
(iv) x =a
(v) y = b
(vi) z = c?
Solution:
(i) The point for which x = 0 is of the form (0, y, z) and lies in yoz plane. Thus required locus of point be yz-plane.

(ii) The point for which y = 0 is of the form (x, 0, z) and lies in the xoz plane.
Thus, required locus of point be xz-plane.

(iii) The point for which z = 0 is of the form (x, y, 0) and lies in xoy plane. Thus required locus of a point be xy-plane.

(iv) Since x = a be the plane parallel to x = 0 i.e. yz plane. Thus locus of a point for which x = a be a plane parallel to yz plane at a distance of a units from it.

(v) Since y = b be the plane parallel to y = 0 i.e. xz-plane. Thus, locus of a point for which y = b be a plane parallel to xz plane at a distance b units from it.

(vi) Since z = c be a plane parallel to plane z = 0 i.e. xy plane. Thus locus of a point for which z = c be a plane || to xy plane at a distance c units from it.

Question 14.
What is the locus of a point for which
(i) x = 0, y = 0
(ii) y = 0, z = 0
(iii) z = 0, x = 0
(iv) x = a, y = b
(v) y = b, z = c
(vi) z = c, x = a?
Solution:
(i) We know that on z-axis, x = 0 = y Thus required locus be z-axis.
(ii) We know that on x-axis, we have y = 0 = z Thus required locus be x-axis.
(iii) We know that on y-axis, we have x = 0 = z Thus, required locus be y-axis.
(iv) x = a be the line || to y-axis and y = b be the line || to x-axis
Thus locus of a point for which x = a, y = b is the line of intersection of given planes x = a and y = b
(v) Clearly the locus of a point for which y = b, z = c is the line of intersection of given planes y = b and z = c
(vi) Given planes are z = c and x = a
Required locus is the line of intersection of planes z = c and x = a.

The availability of step-by-step ISC Mathematics Class 11 OP Malhotra Solutions Chapter 25 Hyperbola Chapter Test can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Chapter Test

Question 1.
Find the eccentricity and the coordinate of foci of the hyperbola 25x² – 9y² = 225.
Solution:
Given eqn. of hyperbola can be written as ;
\(\frac{x^2}{9}\) – \(\frac{y^2}{25}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a² = 9 and b² = 25
We know that b² = a² (e² – 1)
⇒ 25 = 9(e² – 1)
⇒ \(\frac{25}{9}\) + 1 = e²
⇒ e = \(\frac{\sqrt{34}}{3}\) (∵ e > 0)
Thus the required eccentricity of given eqn. (1) be \(\frac{\sqrt{34}}{3}\)
The coordinates of foci are ( ± ae, 0) i.e. \(\left( \pm 3 \times \frac{\sqrt{34}}{3}, 0\right)\) i.e. \(( \pm \sqrt{34}, 0)\).

Question 2.
Find the value (s) of k so that the line 2x + y + k = 0 may touch the hyperbola 3x² – y² = 3.
Solution:
eqn. of given hyperbola be 3x² – y² = 3 ⇒ \(\frac{x^2}{1}\) – \(\frac{y^2}{3}\) = 1
On comparing eqn. (1) with\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a² = 1; b² = 3
eqn. of given line be
2x + y + k = 0 ⇒ y = – 2x – k
We know that the line y = mx + c touches hyperbola (1) if c = ±\(\sqrt{a^2 m^2-b^2}\)
Here m = – 2 and c = – k
Thus, eqn. (2) touches hyperbola (1)
if – k = ± \(\sqrt{1 \times(-2)^2-3}\)
⇒ – k = ± 1
⇒ k = ± 1

Question 3.
From the following information, find the equation of the hyperbola and the equation of the transverse axis.
Focus (-2, 1), Directrix : 2x – 3y + 1 = 0, e = \(\frac{2}{\sqrt{3}}\)
Solution:
Let P (x, y) be any point on the hyperbola.
Then by definition, we have | PF | = | PM |

On squaring both sides ; we have
(x + 2)² + (y – 1)² =\frac{4}{39}(2x – 3y + 1)²
⇒ 39 [x² + 4x + 4 + y² – 2y + 1]
= 4 [4x² + 9y² – 12xy – 6y + 1+ 4x]
which is the required eqn. of hyperbola.
Axis of hyperbola is a line ⊥ to the directrix and pass through the focus (- 2, 1).
Thus eqn. of line ⊥ to 2x – 3y + 1 = 0 be given by
3x + 2y + k = 0 …(1)
eqn. (1) pass through the point (- 2, 1).
-6 + 2 + k = 0
⇒ k = 4
Thus eqn. (1) reduces to ; 3x + 2y + 4 = 0 be the required eqn. of axis of hyperbola.

Question 4.
Find the equation of the hyperbola whose eccentricity is √5 and the sum of whose semi-axes is 9.
Solution:
Let the required eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
given eccentricity of hyperbola be √5
∴ e = √5
Let a be length of semi-major and b be the length of semi-minor axes of hyperbola.
According to given condition, we have
a + b = 9 …(2)
We know that b² = a² (e² – 1)
⇒ b² = a²(5 – 1) = 4a²
⇒ (9 – a)² = 4a² [using eqn. (2)]
⇒ 3a² + 18a – 81 = 0
⇒ (a – 3)(3a + 27) = 0
⇒ a = 3 (∵ a > 0)
∴ from (2); b = 9 – 3 = 6
Thus eqn. (1) reduces to ;
\(\frac{x^2}{9}\) – \(\frac{y^2}{36}\) = 1
which is the required eqn. of hyperbola.

Question 5.
Find the equation of the hyperbola whose foci are (4, 1),(8, 1) and whose eccentricity is 2.
Solution:
We know that centre is the mid-point of line joining the two foci (4, 1) and (8, 1).
∴ Coordinates of centre of hyperbola are \(\left(\frac{4+8}{2}, \frac{1+1}{2}\right)\) i.e. (6, 1).
Since ordinate of both foci are identical
∴ Transverse axis of the hyperbola is parallel to x-axis.
Thus eqn. of hyperbola can be taken as
\(\frac{(x-6)^2}{a^2}\) – \(\frac{(y-1)^2}{b^2}\) = 1
Given eccentricity of eqn. (1) be 2 i.e. e = 2
∴ distance between foci
= \(\sqrt{(8-4)^2+(1-1)^2}\) = 4
⇒ 2ae = 4 ⇒ ae = 2 ⇒ a = 1
∴ b² = a² (e² – 1) = 1²(4 – 1) = 3
Thus, eqn. (1) reduces to ;
\(\frac{(x-6)^2}{1}\) – \(\frac{(y-1)^2}{3}\) = 1
⇒ 3 (x – 6)² – (y – 1)² = 3
which is the required eqn. of hyperbola.

Question 6.
Show that the line y = x + √7 touches the hyperbola 9x² – 16y² = 144.
Solution:
Given eqn. of hyperbola be
9x² – 16y² = 144
and eqn. of given line be
y = x + √7 …(2)
From (2); y = x + √7, putting in eqn. (1); we get
9x² – 16 (x + √7)² = 144
⇒ 9x² – 16 (x² + 7 + 2√7 x) – 144 = 0
⇒ -7x² – 32√7x – 256 = 0
⇒ 7x² + 32√7x + 256 = 0
⇒ (√7x +1 6)² = 0
eqn. (3) is quadratic in x and have equal roots
∴ from (3) ; x = \(\frac{-16}{\sqrt{7}}\)
Thus line (2) touches hyperbola eqn. (1).

Question 7.
Find the equation of the hyperbola whose foci are (0, ± 13) and the length of the conjugate axis is 20.
Solution:
Coordinates of foci are (0, ± 13) which are lies on y-axis and y-axis be the transverse axis of hyperbola.
Let the eqn. of hyperbola be,
\(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1
Here ae = 13
and 2b = 20 and b = 10
We know that, b² = a² (e² – 1)
⇒ 100 = 13² – a²
⇒ a² = 69
Thus eqn. (1) reduces to ; \(\frac{y^2}{69}\) – \(\frac{x^2}{100}\) = 1,
which is the required eqn. of hyperbola.

Question 8.
Find the equation of the hyperbola whose transverse and conjugate axes are the x and y axes respectively, given that the length of conjugate axis is 5 and distance between the foci is 13 .
Solution:
Let the eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
Given length of conjugate axis be 5
∴ 2b = 5 ⇒ b = \(\frac{5}{2}\)
and distance between foci = 13
⇒ 2 ae = 13 …(2)
We know that b² = a² (e² – 1)
⇒ \(\left(\frac{5}{2}\right)^2\) = \(\left(\frac{13}{2}\right)^2\) – a²
⇒ a² = \(\frac{169}{4}\) – \(\frac{25}{4}\) = \(\frac{144}{4}\) = 36
Thus, eqn. (1) reduces to;
\(\frac{x^2}{36}\) – \(\frac{y^2}{25 / 4}\) = 1
⇒ \(\frac{x^2}{36}\) – \(\frac{4 y^2}{25}\) = 1
⇒ 25x² – 144y² = 900
which is the required eqn. of hyperbola.

Question 9.
Find the equation to the conic whose focus is (1, -1), eccentricity is \(\frac{1}{2}\) and the directrix is the line x – y = 3. Is the conic section an ellipse?
Solution:
Given focus of ellipse be F (1, -1) and eqn. of directrix be x – y -3 = 0 and e = \(\frac{1}{2}\)
Let P (x, y) be any point on ellipse. Then by def. of ellipse, we have | PF | = e | PM |
where PM be the ⊥ drawn from P on given directrix.

On squaring both sides, we have
8[(x – 1)² + (y + 1)²] = (x – y – 3)²
⇒ 8 (x² – 2x + y² + 2y + 2) = x² + y² + 9 – 2xy + 6y – 6x
⇒ 7x² + 7y² + 2xy – 10x + 10y + 7 = 0
which is the required eqn. of ellipse.

Utilizing ISC Mathematics Class 11 OP Malhotra Solutions Chapter 25 Hyperbola Ex 25(b) as a study aid can enhance exam preparation.

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Ex 25(b)

Question 1.
Find the tangent to the parabola y² = 16x, making an angle of 45° with the x-axis.
Solution:
The eqn. of tangent to parabola be
y = mx + \(\frac{a}{m}\) …(1)
Here m = tan 45° = 1
On comparing y² = 16x with y² = 4ax ;
we have, 16 = 4a
⇒ a = 4
Thus eqn. (1) reduces to ; y = x + 4
which is the required eqn. of tangent to given parabola.

Question 2.
A tangent to the parabola y² = 16x makes an angle of 60° with the x-axis. Find its point of contact.
Solution:
We know that, the line y = mx + c may touch the parabola y² =4ax then the point of contact be given by \(\left(\frac{a}{m^2}, \frac{2 a}{m}\right)\),
On comparing y² = 16x with y² = 4ax
we have, 16 = 4a ⇒ a = 4
and m = tan 60° = √3
∴ required point of contact be \(\left(\frac{4}{3}, \frac{2 \times 4}{\sqrt{3}}\right)\) i.e. \(\left(\frac{4}{\sqrt{3}}, \frac{8}{\sqrt{3}}\right)\)

Question 3.
(i) Find the equations of the tangents to the parabola y² = 6x which pass through the point \(\left(\frac{3}{2}, 5\right)\).
(ii) Find the equations of the tangents to the parabola y² + 12x = 0 from the point (3, 8).
Solution:
(i) On comparing y² = 6x with y² = 4ax; we have
4a = 6 ⇒ a = \(\frac{3}{2}\)
We know that, eqn. of any tangent to parabola y² = 4ax be given by
y = mx + \(\frac{a}{m}\) …(1)
Thus required eqn. of tangent to parabola y² = 6x be given by
y = mx + \(\frac{3}{2 m}\) …(2)
Now eqn. (2) passes through the point \(\left(\frac{3}{2}, 5\right)\).
5 = \(\frac{3}{2}\) m + \(\frac{3}{2 m}\)
⇒ 10m = 3m² + 3
⇒ 3m² – 10m + 3 = 0
⇒ (m – 3) (3m – 1) = 0
⇒ m = 3, \(\frac{1}{3}\)
putting the values of m in eqn. (2); we have
y = 3x + \(\frac{3}{2 \times 3}\)
⇒ y = 3x + \(\frac{1}{2}\)
⇒ 2y = 6x + 1 …(3)
and y = \(\frac{1}{3}\)x + \(\frac{3 \times 3}{2 \times 1}\)
⇒ y = \(\frac{x}{3}\) + \(\frac{9}{2}\)
⇒ 6y = 2x + 27 …(4)
Thus eqn. (3) and eqn. (4) are the required eqns. of tangents to given parabola.

(ii) Given eqn. of parabola be
y² = -12x …(1)
On comparing eqn. (1) with y² = 4ax
we have, 4a = – 12 ⇒ a = – 3
We know that, eqn. of any tangent to parabola y² = 4ax be given by
y = mx + \(\frac{a}{m}\)
Thus eqn. of any tangent to given parabola (1) be given by
y = mx – \(\frac{3}{m}\)
Now eqn. (2) passes through the point (3, 8).
8 = 3m – \(\frac{3}{m}\) ⇒ 3m² – 8m – 3 = 0
⇒ m = \(\frac{8 \pm \sqrt{64+36}}{6}\) = \(\frac{8 \pm 10}{6}\) = 3, \(\frac{1}{3}\)
putting m = 3 in eqn. (2) ; we have
y = 3x – 1 …(3)
putting m = –\(\frac{1}{3}\) in eqn. (2) ; we have
y = –\(\frac{1}{3}\) + 9 ⇒ 3y = -x + 27 …(4)
Thus, eqn. (3) and eqn. (4) gives the required tangents to given parabola.

Question 4.
Show that the line 12y – 20x – 9 = 0 touches the parabola y² = 5x.
Solution:
Given eqn. of parabola be
y² = 5x …(1)
On comparing eqn. (1) with y² = 4ax ; we have
4a = 5
⇒ a = \(\frac{5}{4}\)
Given eqn. of line can be written as ;
y = \(\frac{20 x}{12}\) + \(\frac{9}{12}\)
⇒ y = \(\frac{5 x}{3}\) + \(\frac{3}{4}\) …(2)
Comparing eqn. (2) with y = mx + c
we have, m = \(\frac{5}{3}\) and c = \(\frac{3}{4}\)
Here, \(\frac{a}{m}\) = \(\frac{\frac{5}{4}}{\frac{5}{3}}\) = \(\frac{3}{4}\) = c
Thus line (2) touches the parabola (1).

Question 5.
Show that the line x + y = 1 touches the parabola y = x – x².
Solution:
Given eqn. of line be
x + y = 1 …(1)
and eqn. of parabola be y = x – x² …(2)
From (1); y = 1 – x, putting in eqn. (2); we have
1 – x = x – x²
⇒ x² – 2x + 1 = 0
⇒ (x – 1)² = 0 …(3)
i.e. eqn. (3) have equal roots.
Thus eqn. (1) touches eqn. (2).

Question 6.
Show that the line x + ny + an² = 0 touches the parabola y² = 4ax and find the point of contact.
Solution:
Given eqn. of line be
x + ny + an² = 0 …(1)
y² = 4ax …(2)
From (1); y = \(\frac{-x-a n^2}{n}\)
putting the value of y in eqn. (2) ; we have

which is quadratic in x and have equal roots. Thus line (1) touches parabola (2).
∴ from (3); x = an²
from (1); ny + 2an² = 0 ⇒ y = – 2an
Hence the required point of contact be (an², – 2an).

Question 7.
Find the tangents to the ellipse x² + 9y² = 3, which are (i) parallel (ii) perpendicular to the line 3x + 4y = 9.
Solution:
Given eqn. of ellipse be
x² + 9y² = 3 ⇒ \(\frac{x^2}{3}\) + \(\frac{y^2}{\frac{1}{3}}\) = 1 …(1)
which is a horizontal ellipse.
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
we have a² = 3 and b² = \(\frac{1}{3}\)
eqn. of given line be
3x + 4y – 9 = 0 …(2)
∴ slope of line (2) = –\(\frac{3}{4}\)
∴ slope of line || to line (2) = –\(\frac{3}{4}\) = m
The eqns. of tangents to ellipse (1) be

(ii) slope of line ⊥ to line (2)
= \(-\frac{1}{-\frac{3}{4}}\) = \(\frac{4}{3}\) = m
∴ required eqns. of tangents to ellipse (1) be given by

Question 8.
Find the equations of the tangents to the ellipse \(\frac{x^2}{2}\) + \(\frac{y^2}{7}\) = 1 that make an angle of 45° with the x-axis.
Solution:
Given eqn. of ellipse be
\(\frac{x^2}{2}\) + \(\frac{y^2}{7}\) = 1 …(1)
which is a vertical ellipse.
On comparing eqn. (1) with
\(\frac{x^2}{b^2}\) + \(\frac{y^2}{a^2}\) = 1, a > b > o
We have a² = 7 ; b² = 2
Here m = tan 45° = 1
The eqns. of tangents to given ellipse (1) be given by
y = mx ± \(\sqrt{b^2 m^2+a^2}\)
⇒ y = x ± \(\sqrt{2+7}\)
⇒ y = x ± 3

Question 9.
Find the equation of the tangents to the ellipse \(\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1, which make equal intercepts on the axes.
Solution:
Given eqn. of ellipse be,
\(\frac{x^2}{16}\) + \(\frac{y^2}{9}\) = 1 …(1)
which is a horizontal ellipse.
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
where a > b > 0
we have a² = 16 and b² = 9
Since tangents makes equal intercepts on axes
∴ slope of tangents = m = ± 1
Thus required eqns. of tangents to eqn. (1) be given by

Question 10.
Find the value of ‘c’ so that 2x -y + c = 0 may touch the ellipse x² + 2y² = 2.
Solution:
Given eqn. of line be 2x -y + c = 0
⇒ y = 2x + c …(1)
and eqn. of given ellipse be x² + 2y² = 2
⇒ \(\frac{x^2}{2}\) + \(\frac{y^2}{1}\) = 1 …(2)
We know that the line y = mx + c touches the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
if c = ± \(\sqrt{a^2 m^2+b^2}\)
if c = ± \(\sqrt{2 \times 2^2+1}\)
⇒ c = ± 3
[Here m = 2 ; a² = 2 ; b² = 1]

Question 11.
Show that the line lx + my = 1 will touch the ellipse \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 if a²l² + b²m² = 1.
Solution:
eqn. of given line be
lx + my = 1 …(1)
and eqn. of ellipse be \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 … (2)
From (1) ; y = \(\frac{1-l x}{m}\)
putting the value of y in eqn. (2); we have
\(\frac{x^2}{a^2}\) + \(\frac{1}{b^2}\) \(\left(\frac{1-l x}{m}\right)^2\) = 1
⇒ m²b²x² + a² (1 – lx)² = a²b²m²
⇒ x² [m²b² + a²l²] – 2a²lx + a² – a²b²m² = 0 …(3)
Now eqn. (1) touches eqn. (2)
if roots of quadratic eqn. (3) are equal
if Discriminant = 0
if (- 2a²l)² – 4 (b²m² + a²l²) (a² – a²b²m²) = 0
if 4a⁴l² – 4a²b²m² + 4a²b⁴m⁴ – 4a⁴l² + 4a⁴b²l²m² = 0
if 4a²b²m² (b²m² + a²l² – 1) = 0
if b²m² + a²l² = 1 which is the required condition.

Question 12.
Show that the following lines are tangents to the given hyperbola and determine the points of contact.
(i) x + 1 = 0, 4x² – 3y² = 4
(ii) x – 2y + 1 = 0, x² – 6y² = 3
Solution:
Given eqn. of line be x + 1 = 0 ⇒ x = – 1 …(1)
and eqn. of given hyperbola be 4x² – 3y² = 4 i.e. \(\frac{x^2}{1}\) – \(\frac{y^2}{\frac{4}{3}}\) = 1
putting eqn. (1) in eqn. (2) ; we have
4 – 3x² = 4 ⇒ y² = 0 which is quadratic in y and gives equal roots and each root be 0.
Thus line (1) touches hyperbola (2).
Putting y = 0 in eqn. (1); x = – 1
Hence the required point of contact be (- 1, 0)

(ii) Given eqn. of line be x- 2y + 1 = 0 …(1)
and eqn. of hyperbola be x² – 6y² = 3 …(2)
From (1); y = \(\frac{x+1}{2}\), putting the value of y in eqn. (2); we have
x² – 6\(\left(\frac{x+1}{2}\right)^2\) = 3
⇒ x² – \(\frac{3}{2}\) (x + 1)² = 3
⇒ 2x² – 3 (x² + 2x + 1) = 6
⇒ -x² – 6x – 9 = 0
⇒ (x – 3)² = 0
Clearly eqn. (3) have equal roots.
Thus line (1) touches given hyperbola (2).
∴ from (3); x + 3 = 0 ⇒ x = – 3
∴ from (1); y = – 1
Thus, the required point of contact be (- 3, – 1).

Question 13.
Find the equations of the tangents to the hyperbola 2x² – 3y² = 6, which re parallel to the Iine x + y – 2 = 0.
Solution:
eqn. of given hyperbola be, 2x² – 3y² = 6 ⇒ \(\frac{x^2}{3}\) – \(\frac{y^2}{2}\) = 1
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have a² = 3 and b² = 2
and eqn. of given line be x + y – 2 = 0
i.e. y = – x + 2 …(2)
On comparing eqn. (2) with y = mx + c
we have m = – 1 and c = 2
Thus the required eqns. of tangents to eqn. (1) be given by
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y =- x ± \(\sqrt{3(-1)^2-2}\)
⇒ y = – x ± 1
⇒ x + y ± 1 = 0

Question 14.
The tangents from P to the hyperbola \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 are mutually perpendicular, show that the locus of P is the circle x² + y² = a² – b².
Solution:
Given eqn. of hyperbola be
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
Thus, eqn. of tangents to hyperbola (1) at P (x, y) be given by
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y – mx = ± \(\sqrt{a^2 m^2-b^2}\)
On squaring both sides ; we have
(y – mx)² = a²m² – b²
⇒ m²x² +y² – 2myx + b² – a²m² = 0
⇒ m²x² – 2myx + (b² – a²m² + y²) = 0
⇒ m²(x² – a²) – 2xym + b² + y² = 0 …(2)
eqn. (2) is a quadratic in m and hence two roots say m₁ and m₂
∴ product of roots = m₁ m₂ = – 1
⇒ \(\frac{y^2+b^2}{x^2-a^2}\) = -1
y² + b² = -x² + a²
⇒ x² +y² = a² – b² which is the required locus of point P.

Question 15.
Show that the straight line x + y = 1 touches the hyperbola 2x² – 3y² = 6. Also find the coordinates of the point of contact.
SoLUTION:
Given eqn. of line be
x + y =1 …(1)
and eqn. of given hyperbola be
2x² – 3y² = 6 …(2)
∴ from (1); y = 1 – x, putting in eqn. (3); we get
2x² – 3 (1 – x)² = 6
⇒ 2x² – 3 (x² + 1 – 2x) = 6
⇒ -x² + 6x – 9 = 0
⇒ x² – 6x + 9 = 0
⇒ (x – 3)² = 0 …(3)
Thus eqn. (3) is a quadratic in x and have two equal roots i.e. x = + 3, + 3.
∴ eqn. (1) touches hyperbola (2).
putting x = + 3 in eqn. (1); we have y = -2.
∴ required point of contact be (+ 3, – 2).

Question 16.
Find the equations of the tangents to the hyperbola 4x² – 9y² = 144, which are perpendicular to the line 6x + 5y = 21. Solution:
Given eqn. of hyperbola be
4x² – 9y² = 144 ⇒ \(\frac{x^2}{36}\) – \(\frac{y^2}{16}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have, a² = 36 and b² = 16
slope of given line 6x + 5y – 21 = 0 be \(\frac{-6}{5}\)
∴ slope of line ⊥ to given line = \(\frac{-1}{-6/5}\) = \(\frac{5}{6}\) = m
Thus, eqns. of tangents to hyperbola (1) be given by
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y = \(\frac{5}{6}\) x ± \(\sqrt{36 \times \frac{25}{36}-16}\)
⇒ y = \(\frac{5}{6}\) x ± 3
⇒ 6y = 5x ± 18
⇒ 5x – 6y ± 18 = 0

Interactive ISC Mathematics Class 11 OP Malhotra Solutions Chapter 25 Hyperbola Ex 25(a) engage students in active learning and exploration.

S Chand Class 11 ICSE Maths Solutions Chapter 25 Hyperbola Ex 25(a)

Question 1.
Find the equation of the hyperbola whose focus is (1, 1), directrix 2x + 2y = 1, and eccentricity √2.
Solution:
Given Focus of hyperbola be F (1, 1) and directrix be 2x + 2y = 1 and e = √2 be the eccentricity of required hyperbola.
Let P (x, y) be any point on hyperbola.
Then by def. | PF | = e | PM |
\(\sqrt{(x-1)^2+(y-1)^2}\) = \(\sqrt{2}\) \(\frac{|2 x+2 y-1|}{\sqrt{2^2+2^2}}\)
= \(\sqrt{(x-1)^2+(y-1)^2}\) = \(\frac { 1 }{ 2 }\) | 2x + 2y – 1 |
On squaring both sides ; we have
4 [(x – 1)² + (y – 1)²] = (2x + 2y – 1)²
⇒ 4 [x² + y² – 2x – 2y + 2]
⇒ 4x² + 4y² + 1 + 8xy – 4y – 4x
⇒ 8xy + 4x + 4y-7 = 0
which is the required eqn. of hyperbola.

Question 2.
Find the equation to the hyperbola whose eccentricity is 2, whose focus is (2, 0) and whose directrix is x – y = 0.
Solution:
Let P (x, y) be any point on hyperbola.
Then by def. | PF | = e | PM |

On squaring both sides ; we have
(x- 2)² + (y²) = 2 (x – y)²
⇒ x² – 4x + 4 + y² = 2(x² + y² – 2xy)
⇒ x² + y² – 4xy + 4x – 4 = 0
which is the required eqn. of hyperbola.

Question 3.
Find the equation to the conic section whose focus is (ae, 0), directrix is the line ex = a, and eccentricity is e. State whether the conic section is an ellipse or a hyperbola.
Solution:
Let P (x, y) be any point on the conic section. Then by def. | PF | = e | PM |

On squaring both sides ; we have
[(x – ae)² + y²] = e²x² + a² – 2aex
x² + a²e² – 2aex + y² = e²x² + a² – 2aex
⇒ x²(1 – e²) + y² = a²
⇒ \(\frac{x^2}{a^2}\) + \(\frac{y^2}{a^2\left(1-e^2\right)}\) = 1
which represents an ellipse if e < 1 and hyperbola if e > 1.

Question 4.
Find the equation of the hyperbola whose axes are along the coordinate axes and which passes through (- 3, 4), and (5, 6).
Solution:
Let the eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
Since it is given that eqn. (1) passes through the point (- 3, 4) and (5, 6).
\(\frac{9}{a^2}\) – \(\frac{16}{b^2}\) = 1 …(2)
and \(\frac{25}{a^2}\) – \(\frac{36}{b^2}\) = 1 …(3)
eqn. (2) × 9 – eqn. (3) × 4; we have
(81 – 100) \(\frac{1}{a^2}\) = 5
⇒ a² = \(\frac{-19}{5}\), which is not possible
Thus we take the eqn. of hyperbola be
\(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1
Now the points (- 3, 4) and (5, 6) lies on eqn. (4); we have
\(\frac{16}{a^2}\) – \(\frac{9}{b^2}\) = 1 …(5)
and \(\frac{36}{a^2}\) – \(\frac{25}{b^2}\) = 1 …(6)
eqn. (5) × 9 – eqn. (6) × 4; we have
(- 81 + 100) \(\frac{1}{b^2}\) = 5 ⇒ b² = \(\frac{19}{5}\)
∴ from (5) ; \(\frac{16}{a^2}\) – \(\frac{9 \times 5}{19}\) = 1
⇒ \(\frac{16}{a^2}\) = 1 + \(\frac{45}{19}\) ⇒ \(\frac{16}{a^2}\) = \(\frac{64}{19}\)
⇒ a² = \(\frac{19}{4}\)
Thus eqn. (4) reduces to ;
\(\frac{4 y^2}{19}\) – \(-\frac{5 x^2}{19}\) = 1
⇒ 4 y² – 5x² = 19
which is the required eqn. of hyperbola.

Question 5.
Find the eccentricity of the hyperbola whose equation is 2x² – 3y² = 15.
Solution:
Given eqn. of hyperbola can be written as ;
\(\frac{x^2}{\frac{15}{2}}\) – \(\frac{y^2}{5}\) = 1
On comparing with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1; we have
a² = \(\frac{15}{2}\) and b² = 5
We know that b² = a² (e² – 1)
⇒ 5 = \(\frac{15}{2}\) (e² – 1) ⇒ e² – 1 = \(\frac{2}{3}\)
⇒ e² = \(\frac{5}{3}\) ⇒ e = \(\sqrt{\frac{5}{3}}\) [∴ e > 0]
Thus required eccentricity of hyperbola be \(\sqrt{\frac{5}{3}}\).

Question 6.
Find the eccentricity and the coordinates of the foci of the curve 3x² – y² = 4.
Solution:
Given eqn. of curve be
3x² – y² = 4 ⇒ \(\frac{x^2}{\frac{4}{3}}\) – \(\frac{y^2}{4}\) = 1
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
we have, a² = \(\frac{4}{3}\) and b² = 4
We know that, b² = a²(e² – 1)
⇒ 4 = \(\frac{4}{3}\) (e² – 1)
⇒ e² = 4
⇒ e = 2 [∵ e > 0]
which gives the required eccentricity of the hyperbola.
Thus coordinates of foci are ( ± ae, 0)
i.e. \(\left( \pm \frac{2}{\sqrt{3}} \times 2,0\right)\) i.e. \(\left( \pm \frac{4}{\sqrt{3}}, 0\right)\).

Question 7.
Find the coordinates of the foci, vertices, eccentricity and the length of the latus rectum of the hyperbola
(i) 16x² – 9y² = 576
(ii) \(\frac{y^2}{9}\) – \(\frac{x^2}{27}\) = 1
(iii) 9y² – 4x² = 36
(iv) 49y² – 16x² = 784
Solution:
(i) Given eqn. be 16x² – 9y² = 576
⇒ \(\frac{x^2}{36}\) – \(\frac{y^2}{54}\) = 1
On comparing with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1
We have a² = 36; b² = 64 i.e. a = 6 ; b = 8
We know that b² = a² (e² – 1)
where e be the eccentricity of the given hyperbola
⇒ 64 = 36 (e² – 1)
⇒ \(\frac{64}{36}\) = e² – 1
⇒ \(\frac{16}{9}\) + 1 = e²
⇒ e² = \(\frac{25}{9}\)
⇒ e = \(\frac{5}{3}\) (∵ e > 0)
The foci of given hyperbola are (± ae , 0)
i.e. \(\left( \pm 6 \times \frac{5}{3}, 0\right)\) i.e. (± 10, 0)
and vertices of given hyperbola are (± a, 0) i.e. (± 6, 0)
length of transverse axis = 2a = 2 × 6 = 12
length of conjugate axis = 2b = 2 × 8 = 16
length of latus-rectum = \(\frac{2 b^2}{a}\) = \(\frac{2 \times 64}{6}\) = \(\frac{64}{3}\)

(ii) eqn. of given hyperbola is \(\frac{y^2}{9}\) – \(\frac{x^2}{27}\) = 1 …(1)
On comparing eqn. (1) with
\(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1; we have
a² = 9 and b² = 27
Let e be the eccentricity of given hyperbola (1).
We know that b² = a² (e² – 1)
⇒ 27 = 9 (e² – 1)
⇒ e = 2 (∵ e > 0)
Coordinates of foci are (0, ± ae)
i.e. (0, ± 3 × 2) i.e. (0, ± 6) and Coordinates of vertices are (0, ± a) i.e. (0, ± 3).
and length of latus rectum = \(\frac{2 b^2}{a}\) = 2 × \(\frac{27}{3}\) = 18

(iii) Given eqn. of hyperbola be
9y² – 4x² = 36
⇒ \(\frac{y^2}{4}\) – \(\frac{x^2}{9}\) = 1
On comparing with \(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1,
We have a² = 4 ⇒ a = 2; b² = 9 ⇒ b = 3
We know that b² = a² (e² – 1)
⇒ 9 = 4 (e² – 1)
⇒ \(\frac{9}{4}\) = e² – 1
⇒ e² = \(\frac{13}{4}\)
⇒ e = \(\frac{\sqrt{13}}{2}\) (∵ e > 0)
The foci of given hyperbola are (0, ± ae)
vertices of given hyperbola are (0, ± a) i.e. (0, ± 2).
length of transverse axis = 2a = 2 × 2 = 4
length of conjugate axis = 2b = 2 × 3 = 6
length of latus-rectum = \(\frac{2 b^2}{a}\) = \(\frac{2 \times 9}{2}\) = 9

(iv) Given eqn. of hyperbola be
49y² – 16x² = 784
⇒ \(\frac{49 y^2}{784}\) – \(\frac{16 x^2}{784}\) = 1
⇒ \(\frac{y^2}{16}\) – \(\frac{x^2}{49}\) = 1
On comparing with \(frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1,
we have a² = 16 ; b² = 49 i.e. a = 4, b = 7
We know that b² = a² (e² – 1) where e be the eccentricity of hyperbola
⇒ 49 = 16(e² – 1) ⇒ \(\frac{49}{16}\) = e² – 1
⇒ e² = \(\frac{65}{16}\) ⇒ e = \(\frac{\sqrt{65}}{4}\) (∵ e > 0)
Thus the foci of given hyperbola are
(0, ± ae) i.e. (0, ± 4 × \( \frac{\sqrt{65}}{4}\))
i.e. (0, ± \( \frac{\sqrt{65}\) )
Vertices of given hyperbola are (0, ± a) i.e. (0, ± 4)
length of transverse axis = 2a = 2 × 4 = 8
length of conjugate axis = 2b = 2 × 7 = 14
∴ length of latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2 \times 49}{4}\) = \(\frac{49}{2}\)

Question 8.
In the hyperbola x² – 4y² = 4, find the length of the axes, the coordinates of the foci, the eccentricity, and the latus rectum, and the equations of the directrices.
Solution:
Given eqn. of hyperbola can be written as ;
\(\frac{x^2}{4}\) – \(\frac{y^2}{1}\) = 1 …(1)
Comparing eqn. (1) with \(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 ;
we have a² = 4 ; b² = 1
Let the eccentricity of eqn. (1) be e.
We know that b² = a²(e² – 1)
⇒ 1 = 4 (e² – 1)
⇒ e² = \(\frac{5}{4} \)
∴ e = \(\frac{\sqrt{5}}{2}\) (∵ e > 0)
∴ length of conjugate axes = 2a = 2 × 2 = 4
length of transverse axes = 2b = 2 × 1 = 2
Thus, coordinates of foci are (± ae, 0)
i.e. \(\left( \pm 2 \times \frac{\sqrt{5}}{2}, 0\right)\) i.e. \(( \pm \sqrt{5}, 0)\)
and length of latus-rectum = \(\frac{2 b^2}{a}\) = \(\frac{2 \times 1}{2}\) = 1
Thus, eqns. of directrices are given by x = ± \(\frac{a}{e}\)
i.e. x = ± \(\frac{2}{\sqrt{5}}\) × 2 ⇒ x = ± \(\frac{4}{\sqrt{5}}\)
⇒ √5x ± 4 = 0

Question 9.
Find (a) the eccentricities, (b) the coordinates of the foci (c) the equations of the directrices of the following hyperbolas
(i) \(\frac{(x-1)^2}{9}\) – \(\frac{y^2}{4}\) = 1
(ii) \(\frac{(x+1)^2}{64}\) – \(\frac{(y-2)^2}{36}\) = 1
Solution:
(i) Given eqn. of hyperbola be,
\(\frac{(x-1)^2}{9}\) – \(\frac{y^2}{4}\) = 1 …(1)
shifting the origin to point (1, 0).
putting x – 1= X and y = Y in eqn. (1); we have
\(\frac{\mathrm{X}^2}{9}\) – \(\frac{\mathrm{Y}^2}{4}\) = 1
On comparing eqn. (2) with
\(\frac{\mathrm{X}^2}{a^2}\) – \(\frac{\mathrm{Y}^2}{b^2}\) = 1
We have, a² = 9 and b² = 4
We know that b² = a² (e² – 1)
⇒ \(\frac{4}{9}\) = e² – 1
⇒ e² = \(\frac{13}{9}\)
⇒ e = \(\frac{\sqrt{13}}{3}\) (∵ e > 0)
The coordinates of foci are (± ae, 0)

and eqns. of directrices be given by
X = ± \(\frac{a}{e}\)
i.e. x – 1 = ± \(\frac{3 \times 3}{\sqrt{13}}\)
be the directrices of eqn. (1).

(ii) Given eqn. of hyperbola be ,
\(\frac{(x+1)^2}{64}\) – \(\frac{(y-2)^2}{36}\) = 1 …(1)
Shift the origin to point (- 1, 2);
putting x + 1 = X and y – 2 – Y in eqn. (1); we have
\(\frac{x^2}{64}\) – \(\frac{Y^2}{36}\) = 1 …(2)
On comparing eqn. (2) with
\(\frac{x^2}{a^2}\) – \(\frac{Y^2}{b^2}\) = 1
we have, a² = 64 and b² = 36
We know that, b² = a² (e² – 1)
⇒ \(\frac{36}{64}\) = e² – 1 ⇒ e² – 1 = \(\frac{9}{16}\)
⇒ e² = \(\frac{25}{16}\)
⇒ e = \(\frac{5}{4}\) [∵ e > 0]
which is the required eccentricity of given hyperbola (1).
Coordinates of foci of eqn. (2) are given by X = ± ae and Y = 0
⇒ x + 1 = ± 8 × \(\frac{5}{4}\) and y – 2 = 0
⇒ x = – 1 ± 10 and 7 = 2
⇒ x = – 11, 9 and 7 = 2
Thus coordinates of foci of eqn. (1) are (-11, 2) and (9, 2).
eqns. of directrices are given by
x = ± \(\frac{a}{e}\) x + 1 = ± \(\frac{8 \times 4}{5}\)
⇒ x = – 1 ± \(\frac{32}{5}\)
⇒ x = \(\frac{- 37}{5}\) and x = \(\frac{27}{5}\)
i.e. 5x + 37 = 0 and 5x – 27 = 0 are the eqns. of directrices of eqn. (1).

Question 10.
Find the equation of the hyperbola, referred to its axes as the axes of coordinates,
(i) whose transverse and conjugate axes are in length respectively 2 and 3 ;
(ii) whose foci are (2, 0) and (- 2, 0) and eccentricity equal to \(\frac{3}{2}\);
(iii) the distance between whose foci is 4 and whose eccentricity is √ 2;
(iv) whose conjugate axis is 3 and the distance between whose foci is 5 ;
(v) vertices are (0, ± 3) and foci (0, ± 5);
(vi) foci are (± 5, 0) and transverse axis of length 8;
(vii) foci are (0, ± 13) and conjugate axis of length 24 ;
(viii) vertices are (± 7, 0) and e = \(\frac{4}{3}\) ;
(ix) foci are (6, 4) and (- 4, 4) and eccentricity is 2.
Solution:
(i) Let the eqn. of hyperbola be
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
given 2a = 2 and 2b = 3
∴ eqn. (1) reduces to r2 v2
\(\frac{x^2}{1}\) – \(\frac{y^2}{\frac{9}{4}}\)
⇒ 9x² – 4y² – 9 = 0
which is the required eqn. of hyperbola,

(ii) Since foci of hyperbola are (± 2, 0) which lies on x-axis.
Thus eqn. of hyperbola can be taken as ;
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
We have clearly found that ae = 2 …(2)
Also given e = \(\frac{3}{2}\) ∴ from (2); a = \(\frac{4}{3}\)
We know that, b² = a² (e² – 1)
⇒ b² = \(\frac{16}{9}\) \(\left[\frac{9}{4}-1\right]\)
= \(\frac{16}{9}\) × \(\frac{5}{4}\) = \(\frac{20}{9}\)
Thus eqn. (1) reduces to ;

which is the required eqn. of hyperbola.

(iii) Let the eqn. of hyperbola be given by
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
given distance between foci = 4
⇒ 2ae = 4 ⇒ ae = 2 …(2)
and given eccentricity of hyperbola be
√2 i.e. e = √2
∴ from (2); a = \(\frac{2}{\sqrt{2}}\) = √2
We know that
b² = a² (e² – 1) = 2 (2 – 1) = 2
Thus eqn. (1) reduces to ;
\(\frac{x^2}{2}\) – \(\frac{y^2}{2}\) = 1
⇒ x² – y² = 2
which is the required eqn. of hyperbola.

(iv) Let the eqn. of hyperbola be,
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
given 2b = 3 ⇒ b = \(\frac{3}{2}\)
and 2ae = 5 ⇒ ae = \(\frac{5}{2}\)
We know that b² = a² (e² – 1)
⇒ \(\frac{9}{4}\) = \(\left(\frac{5}{2}\right)^2\) – a²
⇒ a² = \(\frac{25}{4}\) – \(\frac{9}{4}\) = \(\frac{16}{4}\) = 4
∴ eqn (1) reduces to;
\(\frac{x^2}{4}\) – \(\frac{4y^2}{9}\) = 1
⇒ 9x² – 16y² = 36
which is the required eqn. of hyperbola.

(v) Since the foci of required hyperbola are F (0, – 5) and F’ (0, 5) and both lies on y-axis. Thus transverse axis lie along y-axis. Further, mid point of line segment FF’ be (0, 0). Thus origin be the centre of required hyperbola.
Thus, eqn. of hyperbola can be taken as \(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1 …(1)
Since vertices are given by (0, ± 3)
so a = 3
and foci are given by (0, ± 5)
∴ ae = 5 ⇒ 3e = 5 ⇒ e = \(\frac{5}{3}\) > 1
since b² = a² (e² – 1) = 9 \(\left(\frac{25}{9}-1\right)\)
= 9 \(\left(\frac{16}{9}\right)\) = 16
Thus eqn. (1) reduces to, \(\frac{y^2}{9}\) – \(\frac{x^2}{16}\) = 1
which is the required eqn. of hyperbola,

(vi) Since the foci of required hyperbola are F (5, 0) and F’ (- 5, 0) and both lies on x-axis. Thus transverse axes lies along x-axis. Further mid point of line segment FF’ be (0,0). Thus (0,0) be the centre of hyperbola. Hence the eqn. of hyperbola can be taken as,
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
given length of transverse axis = 8
⇒ 2a = 8 ⇒ a = 4
and foci are (± 5, 0)
∴ ae = 5
⇒ e = \(\frac{5}{4}\)
We know that b² = a² (e² – 1)
∴ b² = 16 \(\left(\frac{25}{16}-1\right)\) = 16 × \(\frac{9}{16}\) = 9
and foci are (± 5, 0) ∴ ae = 5 ∴ e = \(\frac{5}{14}\)
We know that b² = a² (e² – 1)
⇒ b² = 16\(\left(\frac{25}{16}-1\right)\) = 16 × \(\frac{9}{16}\) = 9
Thus eqn. (1) reduces to \(\frac{x^2}{16}\) – \(\frac{y^2}{9}\) = 1
gives the required eqn. of hyperbola.

(vii) Since foci of required hyperbola be (0, ± 13) which lies on y-axis.
Thus, eqn. of hyperbola can be taken as ;
\(\frac{y^2}{a^2}\) – \(\frac{x^2}{b^2}\) = 1 …(1)
On comparing (0, ± 13) with (0, ± ae)
i.e. ae = 13 …(2)
given 2b = 24 ⇒ b = 12
We know that b² = a²(e² – 1)
⇒ 12² = 13² – a²
⇒ a² = 169 – 144 = 25
∴ eqn. (1) reduces to ; \(\frac{y^2}{25}\) – \(\frac{x^2}{144}\) = 1
⇒ 144y² – 25x² = 3600
which is the required eqn. of hyperbola.

(viii) Since the vertices (± 7, 0) of required hyperbola are lies on x-axis.
Let the eqn. of hyperbola be given as under;
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 … (1)
On comparing (± 7,0) with (± a, 0); we
have a = 7 and given e = \(\frac{4}{3}\)
We know that, b² = a²(e² – 1)
⇒ b² = 49\(\left(\frac{16}{9}-1\right)\) = 49 × \(\frac{7}{9}\)
⇒ b² = \(\frac{343}{9}\)
Thus eqn. (1) reduces to ;
\(\frac{x^2}{49}\) – \(\frac{9y^2}{343}\) = 1 ⇒ 7x² – 9y² = 343
which is the required eqn. of hyperbola.

(ix) Given foci are (6, 4) and (- 4, 4).
Let C (α, β) be the centre of hyperbola.
∴ α = \(\frac{6-4}{2}\) = 1; β = \(\frac{4+4}{2}\) = 4
Thus C (1, 4) be the centre of required hyperbola.
Further, ordinates of both foci are identical.
Therefore, transverse axes of hyperbola be parallel to x-axis.
Let the eqn. of hyperbola be given as under:
\(\frac{(x-1)^2}{a^2}\) – \(\frac{(y-4)^2}{b^2}\) = 1 …(1)
∴ Distance between foci
= \(\sqrt{(-4-6)^2+(4-4)^2}\)
= 10 – 2ae
⇒ ae = 5; also given e = 2
⇒ a = \(\frac{5}{2}\)
We know that b² = a² (e² – 1)
b² = \(\frac{25}{4}\) (4 – 1) = \(\frac{75}{4}\)
∴ eqn. (1) reduces to;
\(\frac{4(x-1)^2}{25}\) – \(\frac{4(y-4)^2}{75}\) = 1
⇒ 12 (x – 1)² – 4(y – 4)² = 75
⇒ 12x² – 24x + 12 – 4y² + 32y – 64 = 75
⇒ 12x² – 4y² – 24x + 32y – 127 = 0
which is the required eqn. of hyperbola.

Question 11.
Find the equation of the hyperbola, referred to its axes as the axes of co- ordinates, whose conjugate axis is 4 and which passes through the point (1, – 2). Find also the equation of the conjugate hyperbola.
Solution:
Since the required hyperbola is having centre at origin and transverse axis along x-axis.
Thus the eqn. of hyperbola can be taken as
\(\frac{x^2}{a^2}\) – \(\frac{y^2}{b^2}\) = 1 …(1)
given length of conjugate axis = 2 b = 5
⇒ b = \(\frac{5}{2}\)
Since eqn. (1) passing through the point (1, – 2).
∴ \(\frac{1}{a^2}\) – \(\frac{4}{b^2}\) = 1
⇒ \(\frac{1}{a^2}\) – \(\frac{4 \times 4}{25}\) = 1
⇒ \(\frac{1}{a^2}\) = 1 + \(\frac{16}{25}\) = \(\frac{41}{25}\)
⇒ a² = \(\frac{25}{41}\)
Thus, eqn. (1) reduces to,
\(\frac{41 x^2}{25}\) – \(\frac{4 y^2}{25}\) = 1
⇒ 41x² – 4y² = 25
which is the required eqn. of hyperbola.
Thus the corresponding eqn. of conjugate hyperbola be
\(\frac{41 x^2}{25}\) – \(\frac{4 y^2}{25}\) = 1
⇒ 41x² – 4y² = 25 = 0

Question 12.
Find the locus of a point such that the difference of its distances from (4,0) and (- 4, 0) is always equal to 2. Name the curve.
Solution:
Let P (x, y) be any point on the locus s.t | PF | -1 PF’ | = 2

Let e be the eccentricity of curve (1).
Then b² = a² (e² – 1)
⇒ 15 = e² – 1
⇒ e = 4 > 1 [∵ e > 1]
Thus the required curve represents a hyperbola.

Students appreciate clear and concise ISC Mathematics Class 11 OP Malhotra Solutions Chapter 24 Ellipse Chapter Test  that guide them through exercises.

S Chand Class 11 ICSE Maths Solutions Chapter 24 Ellipse Chapter Test

Question 1.
Find the eccentricity of the ellipse
\(\frac{(x-3)^2}{8}\) + \(\frac{(y-4)^2}{6}\) = 1
Solution:
Eqn. of given ellipse be
\(\frac{(x-3)^2}{8}\) + \(\frac{(y-4)^2}{6}\) = 1 …(1)
which is an eqn. of horizontal ellipse Comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
We have a² = 8 and b² = 6
We know that b² = a² (1 – e²)
⇒ 6 = 8 (1 – e²) ⇒ e² = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)
⇒ e = \(\frac{1}{2}\) (∵ e > 0)
Thus, required eccentricity of given ellipse be \(\frac{1}{2}\)

Question 2.
The distance between the foci of the ellipse 5x² + 9y² = 45 is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
Given eqn. be ellipse can be written as ;
\(\frac{x^2}{9}\) + \(\frac{y^2}{5}\) = 1 …(1)
which is a horizontal ellipse and
On comparing with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1,
where a > b > 0
We have, a² = 9 ; b² = 5
Also, b² = a²(1 – e²)
⇒ \(\frac{5}{9}\) = 1 – e²
⇒ e² = \(\frac{2}{3}\) (∵ e > 0)
Thus required distance between foci = 2ae
= 2 × 3 × \(\frac{2}{3}\) = 4 ∴ Ans. (c)

Question 3.
Find the equation of an ellipse whose latus rectum is 8 and eccentricity is \(\frac{1}{3}\)
Solution:
Given length of latus rectum of an ellipse = \(\frac{2 b^2}{a}\) = 8 ⇒ b² = 4a …(1)
and e = \(\frac{1}{3}\)
We know that b² = a² (1 – e²)
⇒ b² = a² \(\left[1-\frac{1}{9}\right]\) = \(\frac{8}{9}\)a²
⇒ 4a = \(\frac{8}{9}\)a²
⇒ a = \(\frac{36}{8}\) = \(\frac{9}{2}\)
∴ from (1); b² = 4 × \(\frac{9}{2}\) = 18
Thus required eqn. of an ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1; where a > b > 0
i.e. \(\frac{4 x^2}{81}\) + \(\frac{y^2}{18}\) = 1
⇒ 8x² + 9y² = 162

Question 4.
Find the equation of the ellipse whose foci are at (- 2, 4) and (4, 4) and major and minor axis are 10 and 8 respectively. Also, find the eccentricity of the ellipse.
Solution:
Since the foci of required ellipse are F (- 2, 4) and F (4, 4)
∴ distance between foci = 2ae = | FF’ |
= \(\sqrt{(4+2)^2+(4-4)^2}\) = 6
⇒ ae = 3 …(1)
Also length of major axis = 2a= 10 ⇒ a = 5
∴ from (1); e = \(\frac{3}{5}\) which gives the required eccentricity of an ellipse.
Since the y-coordinates of both points F and F’ are equal.
∴ major axis of the required ellipse is parallel to x-axis also the mid point of line segment FF’ gives the centre of required ellipse. Thus centre of ellipse be

Hence, the eqn. of ellipse having centre (1, 4) and major axis is parallel to x-axis is given by
\(\frac{(x-1)^2}{25}\) +\(\frac{(y-4)^2}{16}\) = 1
⇒ 16 (x – 1)² + 25 (y – 4)² = 400
⇒ 16x² + 25y² – 32x – 200y + 16 = 0

Question 5.
Find the equation of the ellipse whose 1 eccentricity is \(\frac { 1 }{ 2 }\) and whose foci are at the points (± 2, 0).
Solution:
Given foci of the ellipse be, (± 2, 0) and lies on x-axis. Thus x-axis be the major axis of an ellipse.
Let us take the eqn. of ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
Here ae = 2 and e = \(\frac { 1 }{ 2 }\)
⇒ a = 4
⇒ a² = 16
We know that b² = a²(1 -e²)
⇒ b² = 16 \(\left(1-\frac{1}{4}\right)\) = 16 × \(\frac { 3 }{ 4 }\) = 12
Thus eqn. (1) reduces to ; \(\frac{x^2}{16}\) + \(\frac{y^2}{12}\) = 1
which is the required eqn. of an ellipse.

Question 6.
Find the equation of the ellipse whose centre is the origin, major axis \(\frac { 9 }{ 2 }\) and eccentricity \(\frac{1}{\sqrt{3}}\) where the major axis is the horizontal axis.
Solution:
Since the major axis is the horizontal axis and let the eqn. of ellipse can be taken as
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
given 2a = \(\frac { 9 }{ 2 }\) ⇒ a = \(\frac { 9 }{ 4 }\) and e = \(\frac{1}{\sqrt{3}}\)
We know that b² = a² (1 – e²)
⇒ b² = \(\left(\frac{9}{4}\right)^2\) \(\left[1-\frac{1}{3}\right]\) = \(\frac{81}{16}\) × \(\frac{2}{3}\) = \(\frac{27}{8}\)
Thus eqn. (1) reduces to ;
\(\frac{16 x^2}{81}\) + \(\frac{8 y^2}{27}\) = 1
⇒ 16x² + 24y² = 81
which is the required eqn. of an ellipse.

Question 7.
Find the equation of the ellipse whose minor axis is 4 and which has a distance of 6 units between foci.
Solution:
Let us consider the eqn. of ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1; where a > b > 0 …(1)
given 2b = 4 ⇒ b = 2
Also distance between foci = 6
⇒ 2ae = 6 ⇒ ae = 3 …(2)
We know that b² = a² (1 – e²)
⇒ 4 = a² – 9 [using eqn. (2)]
⇒ a² = 13
Thus eqn. (1) reduces to; \(\frac{x^2}{13}\) + \(\frac{y^2}{4}\) = 1
which is the required eqn. of an ellipse.

Question 8.
Find the equation of the ellipse with its centre at (4, – 1), focus at (1, – 1) and given that i passes through (8, 0).
Solution:
Let S’ (α, β) be the other foci of the required ellipse. Thus C (4, – 1) be the mid-point of SS’.
∴ \(\frac{\alpha+1}{2}\) = 4 ⇒ α = 7
and \(\frac{\beta-1}{2}\) = – 1 ⇒ β = – 1
Thus the coordinates of other foci are S’ (7,-1).
Since the ordinate of points S and S’ are equal. Thus major axis is parallel to x-axis and it is a horizontal ellipse.
Let the eqn. of ellipse be
\(\frac{(x-4)^2}{a^2}\) + \(\frac{(y+1)^2}{b^2}\) = 1 …(1)
where a > b > 0
Distance between foci = 2ae = 6 ⇒ ae = 3
We know that b² – a² (1 – e²) = a² – 9
Thus eqn. (1) reduces to ;
\(\frac{(x-4)^2}{a^2}\) + \(\frac{(y+1)^2}{a^2-9}\) = 1 …(2)
eqn. (2) passes through the point (8, 0),
We get \(\frac{(8-4)^2}{a^2}\) + \(\frac{(y+1)^2}{a^2-9}\) = 1
⇒ \(\frac{16}{a^2}\) + \(\frac{1}{a^2-9}\) = 1
⇒ 16 (a² – 9) + a² = a² (a² – 9)
⇒ 17a² – 144 = a⁴ – 9a²
⇒ a⁴ – 26a² + 144 = 0
⇒ a⁴ – 18a² – 8a² + 144 = 0
⇒ a² (a² – 18) – 8 (a² – 18) = 0
⇒ (a² – 18) (a² – 8) = 0
⇒ a² = 18, 8
When a² = 8
∴ b² = a² – 9 = 8 – 9 = – 1
which is false
Thus a² = 18
∴ eqn. (2) reduces, to ;

Question 9.
Find the coordinates of the vertices and the foci and the length of the latus rectum of the ellipse 9x² + 25y² = 225.
Solution:
Given eqn. of ellipse can be written as
\(\frac{x^2}{25}\) + \(\frac{y^2}{9}\) = 1 …(7)
which is a horizontal ellipse.
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1,
where a > b > 0.
we have, a² = 25 and b² = 9
i.e. a = 5 and 6 = 3 (∵ a, b > 0)
Thus the coordinates of vertices are (± ae, 0) i.e. (± 5, 0).
We know that, b² = a² (1 – e²)
⇒ 9 = 25 (1 – e²) ⇒ 1 – e² = \(\frac{9}{25}\)
⇒ e² = \(\frac{16}{25}\) ⇒ e² = \(\frac{4}{5}\) (∵ e > 0)
Coordinates of foci are (± ae, 0)

and length of latus rectum = \(\frac{2 b^2}{a}\) = 2 × \(\frac{9}{5}\) = \(\frac{18}{5}\) units

Question 10.
Find the eccentricity and the equations of the directrices of the ellipse 7x² + 16y² = 112.
Solution:
Given eqn. of ellipse can be written as ;
\(\frac{x^2}{16}\) + \(\frac{y^2}{7}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
We have, a² = 16 b² = 7
We know that, b² = a² (1 – e²)
⇒ \(\frac{7}{16}\) = 1 – e² ⇒ e² = 1 – \(\frac{7}{16}\) = \(\frac{9}{16}\)
⇒ e = \(\frac{3}{4}\)
Thus required eqns. of directrices are given
by x = ± \(\frac{a}{e}\) = ± \(\frac{4 \times 4}{3}\) ⇒ 3x ± 16 = 0

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S Chand Class 11 ICSE Maths Solutions Chapter 24 Ellipse Ex 24

Question 1.
Find the eccentricity of the ellipse of which the major axis is double the minor axis.
Solution:
Let a be the length of semi-major and semi-minor axis of the ellipse.
According to given condition, a = 2b
We know that b² = a² (1 – e²)
⇒ b² = 4b²(1 – e²)
⇒ \(\frac { 1 }{ 4 }\) = 1 – e² ⇒ e² = 1 – \(\frac { 1 }{ 4 }\) = \(\frac { 3 }{ 4 }\)
⇒ e = \(\frac{\sqrt{3}}{2}\) [∵ e > 0]
Thus required eccentricity of an ellipse be \(\frac{\sqrt{3}}{2}\).

Question 2.
If the minor axis of an ellipse is equal to the distance between its foci, prove that its eccentricity is \(\frac{1}{\sqrt{2}}\).
Solution:
Let e be the eccentricity of an ellipse according to given condition, we have
2b = 2ae ⇒ b = ae
We know that b² = a² (1 – e²)
⇒ a²e² = a² (1 – e²)
⇒ e² = 1 – e²
⇒ 2e² = 1 ⇒ e = \(\frac{1}{\sqrt{2}}\) [∵ e > 0]

Question 3.
Find the latus rectum and eccentricity of the ellipse whose semi-axes are 5 and 4.
Solution:
Given a = 5 and b = 4
Let e be the eccentricity of an ellipse Then b² – a² (1 – e²)
⇒ 16 = 25 (1 – e²)
⇒ 1 – e² = \(\frac { 16 }{ 25 }\) ⇒ e² = \(\frac { 9 }{ 25 }\)
⇒ e = \(\frac { 3 }{ 5 }\) (∵ e > 0)
∴ length of latus-rectum = \(\frac{2 b^2}{a}\)
= 2 × \(\frac{4^2}{5}\) = \(\frac { 32 }{ 5 }\)

Question 4.
Find the eccentricity of the ellipse whose latus rectum is (/) half its major axis, (ii) half its minor axis.
Solution:
Let a be the length of semi-major and b be the length of semi-minor axis of an ellipse and e be the eccentricity of an ellipse.
(i) According to given condition,

Question 5.
If the eccentricity is zero, prove that the ellipse becomes a circle.
Solution:
We know that, b² = a² (1 – e²) …(1)
We have given e = 0 ∴ from (1) ; we have
b² = a² b = a (∵ b, a > 0)
Thus the given eqn. of ellipse reduces to,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{a^2}\) = 1
⇒ x² + y² = a²

Question 6.
Find the equation to the ellipse with axes as the axes of coordinates.
(i) major axis = – 6, minor axis = 4 ;
(ii) which passes through the points (- 3, 1) and (2, – 2) ;
(iii) axes are 10 and 8 and the major axis along
(a) the axis of x,
(b) the axis of y ;
(iv) major axis \(\frac{9}{2}\) and eccentricity \(\frac{1}{\sqrt{3}}\), where the major axis is the horizontal axis ;
(v) latus rectum is 5 and eccentricity \(\frac{2}{3}\),
(vi) foci are (± 4, 0) and e = \(\frac{1}{3}\);
(vii) distance between the foci is 10 and its latus rectum is 15 ;
(viii) distance of the focus from the corresponding directrix is 9 and eccentricity is \(\frac{4}{5}\);
(ix) the minor axis is equal to the distance between the foci, and the latus rectum is 10.
Solution:
(i) Let a and b are the lengths of major and minor axes of an ellipse respectively and the eqn. of ellipse be
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
given 2a = 6 ⇒ a = 3
and 2b = 4 ⇒ b = 2
∴ eqn. (1) reduces to;
\(\frac{x^2}{9}\) + \(\frac{y^2}{4}\) = 1
⇒ 4x² + 9y² = 36

(ii) Let the eqn. of ellipse be
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1, where a > b > 0 …(1)
eqn. (1) pass through the points (- 3, 1) and (2, – 2)
∴ \(\frac{9}{a^2}\) + \(\frac{1}{b^2}\) = 1 …(2)
\(\frac{4}{a^2}\) + \(\frac{4}{b^2}\) = 1 …(2)
4 × eqn. (2) – eqn. (3); we have
\(\frac{32}{a^2}\) = 3 ⇒ a² = \(\frac{32}{3}\)
∴ from (2); \(\frac{9 \times 3}{32}\) + \(\frac{1}{b^2}\) = 1
⇒ \(\frac{1}{b^2}\) = \(\frac{5}{32}\)
⇒ b² = \(\frac{32}{5}\)
Thus eqn. (2) reduces to;
\(\frac{3 x^2}{32}\) + \(\frac{5 y^2}{32}\) = 1;
⇒ 3x² + 5y² = 32

(iii) (a) Here, major axis along x-axis and minor axis along y-axis. Let the eqn. of ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1, where a > b > 0 …(1)
given 2a = 10 ⇒a = 5
and 2b = 8 ⇒ b = 4
∴ eqn. (1) reduces to;
\(\frac{x^2}{25}\) + \(\frac{y^2}{16}\) = 1 ⇒ 16x² + 25y² = 400

(b) Here, major axes along y-axis and minor axis along x-axis.
Let the eqn. of ellipse be taken as
\(\frac{x^2}{b^2}\) + \(\frac{y^2}{a^2}\) = 1 …(1)
when a> b> 0
according to given a = 5 and b = 4
∴eqn. (1) reduces to ; \(\frac{x^2}{16}\) + \(\frac{y^2}{25}\) = 1
⇒ 25x² + 16y² = 400
which is the required eqn. of an ellipse,

(iv) Since the major axis is the horizontal axis and let the eqn. of ellipse can be taken as
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
wher a > b > 0
given 2a = \(\frac{9}{2}\) ⇒ a = \(\frac{9}{4}\) and e = \(\frac{1}{\sqrt{3}}\)
We know that b² = a² (1 – e² )
⇒ b² = \(\left(\frac{9}{4}\right)^2\left[1-\frac{1}{3}\right]\) = \(\frac{81}{16}\) × \(\frac{2}{3}\) = \(\frac{27}{8}\)
Thus eqn. (1) redues to ;
\(\frac{16 x^2}{81}\) + \(\frac{8 y^2}{27}\) = 1 ⇒ 16x² + 24y² = 81
which is the required eqn. of an ellipse,

(v) Let a and b are the lengths of semi-major and semi-minor axes of an ellipse. Let e be the eccentricity of an ellipse. Let the eqn. of ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
given e = \(\frac{2}{3}\) and \(\frac{2 b^2}{a}\) = 5
⇒ b² = \(\frac{5 a}{2}\)
We know that, b² = a²(1 – e²)
⇒ \(\frac{5 a}{2}\) = a² \(\left(1-\frac{4}{9}\right)\) = \(\frac{5 a^2}{9}\)
⇒ a = \(\frac{9}{2}\)
∴ from (1) ; b² = \(\frac{5}{2}\) × \(\frac{9}{2}\) = \(\frac{45}{4}\)
Putting the value of a and b in eqn. (1); we have
\(\frac{4 x^2}{81}\) + \(\frac{4 y^2}{45}\) = 1
⇒ \(\frac{20 x^2+36 y^2}{405}\) = 1
⇒ 20x² + 36y² = 405
Which is the required eqn. of an ellipse.

(vi) given foci are (± 4, 0) i.e. both foci lies on x-axis and hence x-axis be the major axes.
Thus eqn. of ellipse can be written as \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
Its foci are (± ae, 0) ∴ ae = 4
and also given e = \(\frac{1}{3}\)
∴ a = 12
We know that b² = a² (1 – e²)
= 144 \(\left(1-\frac{1}{9}\right)\) = 144 × \(\frac{8}{9}\)
⇒ b² = 128
Thus, eqn.(1) reduces to;
\(\frac{x^2}{144}\) + \(\frac{y^2}{128}\) = 1 ⇒ 8x² + 9y² – 1152 = 0
which is the required eqn. of an ellipse.

(vii) Let a and b be the length of semi-major and semi-minor axes of an ellipse and let the eqn. of ellipse can be taken as
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
given distance between foci = 10
⇒ 2ae = 10 ⇒ ae = 5 …(2)
and \(\frac{2 b^2}{a}\) = 15 ⇒ b² = \(\frac{15a}{2}\) …(3)
We know that b² = a² (1 – e²)
⇒ \(\frac{15a}{2}\) = a² – 25
⇒ 2a² – 15a – 50 = 0
⇒ 2a² – 20a + 5a – 50 = 0
⇒ 2a (a – 10) + 5 (a – 10) = 0
⇒ (a-10) (2a + 5) = 0
⇒ a= 10 (∵ a > 0)
∴ from (3); b² = \(\frac{15 \times 10}{2}\) = 75
Thus eqn. (1) reduces to ;
\(\frac{x^2}{100}\) + \(\frac{y^2}{75}\) = 1
⇒ 3x² + 4y² = 300
which is the required eqn. of an ellipse.

(viii) Given distance of focus (ae, 0) from
directrix x – \(\frac{a}{e}\) = 0 = 9 (given)

(ix) Let a be the length of semi-major axis and b be the length of semi-minor axis and let e be the eccentricity of required ellipse.
Let the required eqn. of an ellipse be,
\(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 …(1)
where a > b > 0
According to given condition,
2b = 2 ae ⇒ b = ae …(2)
and \(\frac{2 b^2}{a}\) = 10
⇒ b² = 5a …(3)
We know that, b² = a² (1 – e²)
⇒ b² = a² – b²
⇒ 2b² = a²
⇒ 2 × 5a = a² [using (2) and (3)]
⇒ a = 10
∴ b² = 5 × 10 = 50
Thus eqn. (1) reduces to ;
\(\frac{x^2}{100}\) + \(\frac{y^2}{50}\) = 1
⇒ x² + 2y² = 100
which is the required eqn. of an ellipse.

Question 7.
Find the equation of the ellipse whose centre is at (- 2, 3) and whose semi-axes are 3 and 2, when the major axis is
(i) parallel to the axes of x ;
(ii) parallel to the axis of y.
Solution:
(i) Since the centre of required ellipse be C (- 2, 3) and mojor axis is parallel to x-axis.
Hence the eqn. of ellipse can be taken as,
\(\frac{(x+2)^2}{9}\) + \(\frac{(y-3)^2}{b^2}\) = 1 …(1)
where a > b > 0
Clearly length of semi-major axis = a = 3
and length of semi-minor axis = b = 2
∴ eqn. (1) reduces to,
\(\frac{(x+2)^2}{9}\) + \(\frac{(y-3)^2}{4}\) = 1
⇒ 4 (x² + 4x + 4) + 9 (y² – 6y + 9) = 36
⇒ 4x² + 9y² + 16x – 54y + 61 = 0
which is the required eqn. of ellipse.

(ii) The eqn. of ellipse having centre C (-2, 3) and major axis is parallel to y-axis be taken as
\(\frac{(x+2)^2}{b^2}\) + \(\frac{(y-3)^2}{a^2}\) = 1 …(1)
where a > b > 0
given length of semi-major axis = a = 3
and length of semi-minor axis = b = 2
∴ eqn. (1) reduces to,
\(\frac{(x+2)^2}{4}\) + \(\frac{(y-3)^2}{9}\) = 1
⇒ 9 (x + 2)² + 4(y – 3)² = 36
⇒ 9x² + 4y² + 36x – 24y + 36 = 0
which is the required eqn. of an ellipse.

Question 8.
Find the equation of the ellipse with its centre at (4, – 1), focus at (1, – 1), and passing through (8, 0).
Solution:
Let S’ (α, β) be the other foci of the required ellipse. Thus C (4, -1) be the mid-point of SS’.

Thus the coordinates of other foci are S’ (7, – 1).
Since the ordinate of points S and S’ are equal. Thus major axis is parallel to x-axis and it is a horizontal ellipse.
Let the eqn. of ellipse be
\(\frac{(x-4)^2}{a^2}\) + \(\frac{(y+1)^2}{b^2}\) = 1 …(1)
where a > b > 0
Distance between foci = 2ae = 6 ⇒ ae = 3
We know that b² = a² (1 – e²) = a² – 9
Thus eqn. (1) reduces to;
\(\frac{(x-4)^2}{a^2}\) + \(\frac{(y+1)^2}{a^2-9}\) = 1 …(2)
eqn. (2) passes through the point (8, 0), we get
\(\frac{(8-4)^2}{a^2}\) + \(\frac{1}{a^2-9}\) = 1
⇒ \(\frac{16}{a^2}\) + \(\frac{1}{a^2-9}\) = 1
⇒ 16 (a² – 9) + a² = a² (a² – 9)
⇒ 17a² – 144 = a⁴ – 9a²
⇒ a⁴ – 26a² + 144 = 0
⇒ a⁴ – 18a² – 8a² + 144 = 0
⇒ a² (a² – 18)- 8 (a² – 18) = 0
⇒ (a² – 18) (a² – 8) = 0
⇒ a² =18, 8
When a² = 8 ∴ b² = a² – 9 = 8 – 9 = – 1
which is false
Thus a² = 18 ∴ eqn. (2) reduces to ;
\(\frac{(x-4)^2}{18}\) + \(\frac{(y+1)^2}{18-9}\) = 1
⇒ \(\frac{(x-4)^2}{18}\) + \(\frac{(y+1)^2}{9}\) = 1
⇒ (x – 4)² + 2 (y + 1)² = 18
⇒ x² + 2y² – 8x +4y = 0
which is the required eqn. of ellipse.

Question 9.
Find the equation of the ellipse with its centre at (3, 1), vertex at (3, – 2), and eccentricity equal to \(\frac{1}{3}\) .
Solution:
Since the centre of required ellipse be C (3, 1) and vertex A (3, – 2). Let the other vertex of ellipse be A’ (α, β). Then C (3, 1) be the mid point of line segment A A’.
∴ 3 = \(\frac{\alpha+3}{2}\) ⇒ α = 3
and 1 = \(\frac{\beta-2}{2}\) ⇒ β = 3
∴ Coordinates of other vertex of an ellipse be A’ (3, 4).
Also x-coordinates of A and A’ are equal
∴ major axis of required ellipse is parallel to y-axis.
Now a = | CA | = | CA’ | = 3
and e = eccentricity of ellipse = \(\frac{1}{3}\)
∴ b² = a² (1 – e)² = 9 \(\left(1-\frac{1}{9}\right)\) = \(\frac{9 \times 8}{9}\) = 8
Thus required eqn. of ellipse having centre C (3, 1) and major axis parallel toy-axis is given by
\(\frac{(x-3)^2}{b^2}\) + \(\frac{(y-1)^2}{a^2}\) = 1
⇒ \(\frac{(x-3)^2}{8}\) + \(\frac{(y-1)^2}{9}\) = 1
⇒ 9 (x – 3)²+ 8(y – 1)² = 72
⇒ 9x² + 8y² – 54x – 16y + 17 = 0

Question 10.
Find the equation of the ellipse whose centre is at (0, 2) and major axis along the axis of y and whose minor axis is equal to the distance between the foci and whose latus rectum is 2.
Solution:
Let the eqn. of ellipse with centre at (0, 2) and having major axis along y-axis be given by
\(\frac{(x-0)^2}{b^2}\) + \(\frac{(y-2)^2}{a^2}\) = 1 …(1)
where a > b > 0
Also, 2b = 2ae ⇒ b = ae …(2)
Further \(\frac{2 b^2}{a}\) = 2 ⇒ b² = a …(3)
We know that, b² = a² (1 – e²)
⇒ a = a² – a (using (2) and (3)]
⇒ 2a = a² ⇒ a = 2 [∵ a > 0]
from (3); b² = 2
Thus eqn. (1) reduces to ;
\(\frac{x^2}{2}\) + \(\frac{(y-2)^2}{4}\) = 1
⇒ 2x² + (y – 2)² = 4
⇒ 2x² + y² – 4y = 0
which is the required eqn. of an ellipse.

Question 11.
Find the equation of the ellipse with (i) focus at (1, – 1), directrix x = 0, and e = \(\frac{\sqrt{2}}{2}\);
(ii) focus at (0, 0), eccentricity is \(\frac{5}{6}\), and directrix is 3x + 4y – 1 = 0.
Solution:
(i) Let P (x, y) be any point on the parabola s.t | PF | = e | PM |
\(\sqrt{(x-1)^2+(y+1)^2}\) = \(\frac{\sqrt{2}|x|}{2}\)
On squaring both sides ; we have
(x – 1)² + (y + 1)² = \(\frac { 1 }{ 2 }\)x²
⇒ 2 [(x – 1)² + (y + 1)²] = x²
⇒ x² + 2y² – 4x + 4y + 4 = 0
which is the required eqn. of an ellipse.

(ii) Let P (x, y) be any point on ellipse
Then by def. | PF | = e | PM |

On squaring both sides ; we have
36(x² + y²) = (3x + 4y – 1)²
⇒ 36 x² + 36 y² = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 27x² + 20y2 – 24xy + 8y + 6x – 1 = 0
which is the required eqn. of an ellipse.

Question 12.
Find the equation of the ellipse from the following data : axis is coincident with x = 1, centre is (1, 5), focus is (1, 8) and the sum of the focal distances of a point on the ellipse is 12.
Solution:
Let the eqn. of ellipse be
\(\frac{(x-1)^2}{b^2}\) + \(\frac{(y-5)^2}{a^2}\) = 1 …(1)
Since axis of ellipse is coincident with x = 1 i.e. required ellipse be a vertical ellipse, given centre of ellipse be C (1, 5) and Focus be S (1, 8).
∴ | CS | = \(\sqrt{(1-1)^2+(8-5)^2}\) = 3
⇒ ae = 3 …(2)
Further sum of focal distances from any point on the ellipse = |PS| + |PS’| = 2a = 12
⇒ a = 6
∴ from (2); e = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
We know that b² = a² (1 – e²)
⇒ b² = 36\(\left(1-\frac{1}{4}\right)\)
⇒ b² = 36 × \(\frac { 3 }{ 4 }\) = 27
∴ eqn. (1) reduces to ;
\(\frac{(x-1)^2}{27}\) + \(\frac{(y-5)^2}{36}\) = 1
which is the required eqn. of an ellipse.

Question 13.
A point P (x, y) moves so that the product of the slopes of the two lines joining P to the two points (- 2, 1) and (6, 5) is – 4. Show that the locus is an ellipse and locate its centre.
Solution:
Let the given points are A (- 2, 1) and B (6, 5).
slope of line joining P (x, y) and A (- 2, 1) = \(\frac{y-1}{x+2}\) = m₁
slope of line joining P (x, y) and B (6, 5) = \(\frac{y-5}{x-6}\) = m₂
product of slopes of both lines = – 4
⇒ \(\left(\frac{y-1}{x+2}\right)\) \(\left(\frac{y-5}{x-6}\right)\) = – 4
⇒ (y – 1) (y – 5) + 4 (x + 2) (x – 6) = 0
⇒ y² + 4x² – 6y – 16y – 43 = 0
which is the required eqn. of an ellipse.
4x² – 16x + y² – 6y = 43
⇒ 4 (x² – 4x + 4 – 4) + (y² – 6y + 9 – 9) = 43
⇒ 4 (x – 2)² + (y – 3)² = 68
⇒ \(\frac{(x-2)^2}{17}\) + \(\frac{(y-3)^2}{68}\) = 1
Clearly centre of an ellipse be C (2, 3).

Question 14.
Find the eccentricity, the coordinates of the foci, and the length of the latus rectum of the ellipse 2x² + 3y² = 1.
Solution:
Given eqn. of an ellipse be
2x² + 3y² = 1
⇒ \(\frac{x^2}{1 / 2}\) + \(\frac{y^2}{1 / 3}\) = 1 …(1)
On comparing with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1,
where a > b > 0
a² = \(\frac{1}{2}\); b² = \(\frac{1}{3}\)
We know that b² = a² (1 – e²)
⇒ \(\frac{1}{3}\) = \(\frac{1}{2}\) (1 – e²) ⇒ \(\frac{2}{3}\) = 1 – e²
⇒ e² = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
⇒ e = \(\frac{1}{\sqrt{3}}\) [∵ e > 0]
Thus required eccentricity of an ellipse be \(\frac{1}{\sqrt{3}}\). Foci are given by (± ae, 0)

Question 15.
For the ellipse, 9x² + 16y² = 576, find the semi-major axis, the semi-minor axis, the eccentricity, the coordinates of the foci, the equations of the directrices, and the length of the latus rectum.
Solution:
Given eqn. of an ellipse be
9x² + 16y² = 576
⇒ \(\frac{x^2}{64}\) + \(\frac{x^2}{64}\) = 1 ….(1)
On comparing with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1;
where a > b > 0
∴ a² = 64 and b² = 36
We know that b² = a² (1 – e²)
⇒ 36 = 64 (1 – e²)
⇒ \(\frac{36}{64}\) = 1 – e²
⇒ e² = 1 – \(\frac{36}{64}\) = 1 – \(\frac{9}{16}\) = \(\frac{7}{16}\)
∴ e = \(\frac{\sqrt{7}}{4}\) [∵ e > 0]
Thus required eccentricity of an ellipse be \(\frac{\sqrt{7}}{4}\)
Here, a = 8 ; b = 6 (∵ a > b > 0)
∴ length of semi major-axis = a = 8
and length of semi minor axis = b = 6
foci are given by (± ae, 0)
i.e. \(\left( \pm 8 \times \frac{\sqrt{7}}{4}, 0\right)\) i.e. \(( \pm 2 \sqrt{7}, 0)\)
length of latus rectum = \(\frac{2 b^2}{a}\) = \(\frac{2 \times 36}{8}\) = 9
eqns. of directrices are given by x = ±\(\frac{a}{e}\)
i.e. x = ±\(\frac{8}{\sqrt{7}}\) × 4 = ±\(\frac{32}{\sqrt{7}}\).

Question 16.
Find the length of the axes, the co-ordinates of the foci, the eccentricity, and latus rectum of the ellipse 3x² + 2y² = 24.
Solution:
Given eqn. of ellipse be 3x² + 2y² = 24
⇒ \(\frac{x^2}{8}\) + \(\frac{y^2}{12}\) = 1;
where a > b > 0
On comparing with \(\frac{x^2}{b^2}\) + \(\frac{y^2}{a^2}\) = 1
we have a² = 12 ; b² = 8
i.e. a = 2√3
and b = 2√2
length of major axis = 2a = 4√3
and length of minor axis = 2b = 4√2
we know that b2 = a² (1 – e²)
⇒ 8 = 12(1 – e²)
⇒ 1 – e² = \(\frac{2}{3}\)
⇒ e² = \(\frac{1}{3}\)
⇒ e = \(\frac{1}{\sqrt{3}}\) (∵e > 0)
Thus foci of an ellipse be (0, ± ae)
i.e. \(\left(0, \pm 2 \sqrt{3} \times \frac{1}{\sqrt{3}}\right)\) i.e. (0, ± 2)
length of latus-rectum = \(\frac{2 b^2}{a}\) = \(\frac{2 \times(2 \sqrt{2})^2}{2 \sqrt{3}}\) = \(\frac{8}{\sqrt{3}}\)

Question 17.
Find the eccentricity of the ellipse, 4x² + 9y² – 8x – 36y + 4 = 0.
Solution:
Given eqn. of an ellipse be,
4x² + 9y² – 8x – 36y + 4 = 0
⇒ 4(x² – 2x + 1 – 1) + 9(y² – 4y + 4 – 4) + 4 = 0
⇒ 4 [(x – 1)² – 1] + 9 [y – 2)² – 4] + 4 = 0
⇒ 4 (x – 1)² + 9 (y – 2)² = 36
⇒ \(\frac{(x-1)^2}{9}\) + \(\frac{(y-2)^2}{4}\) = 1 …(1)

On comparing with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 ; where a > b > 0
Here, a² = 9 and b² = 4
we know that, b² = a² (1 – e²) ⇒ 4 = 9 (1 – e²) ⇒ 1 – e² = \(\frac{4}{9}\)
⇒ e² = \(\frac{5}{9}\) ⇒ e = \(\frac{\sqrt{5}}{3}\) (∵ e > 0)
Thus, required eccentricity of an ellipse be \(\frac{\sqrt{5}}{3}\).

Question 18.
find the centre of the ellipse, \(\frac{x^2-a x}{a^2}\) + \(\frac{y^2-b y}{b^2}\) = 0.
Solution:
Given eqn. of an ellipse be

Question 19.
Find the distance between a focus and an extremity of the minor axis of the ellipse
(i) 4x² + 5y² = 100
(ii) \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
Solution:
(i) Given eqn. of ellipse be 4x² + 5y² = 100
⇒ \(\frac{x^2}{25}\) + \(\frac{y^2}{20}\) = 1 …(1)
On comparing eqn. (1) with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
we have
where a > b > 0
a² = 25 and b² = 20
We know that b² = a² (1 – e²)
⇒ 20 = 25 (1 – e²)
⇒ \(\frac{4}{5}\) = 1 – e²
⇒ e² = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
⇒ e = \(\frac{1}{\sqrt{5}}\) (∵ e > 0)
∴ required distance = distance between focus (ae, 0) and (0, b)

(ii) Given eqn. of ellipse be \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
∴ required distance = \(\sqrt{b^2+a^2 e^2}\) = \(\sqrt{a^2}\) = a
[∵ b² = a² (1 – e)² ⇒ b² + a²e² = a²]

Question 20.
Given the ellipse 36x² + 100y² = 3600, find the equation and the lengths of the focal radii drawn through the point \(\left(8, \frac{18}{5}\right)\).
Solution:
Given eqn. of ellipse be,
\(\frac{x^2}{100}\) + \(\frac{y^2}{36}\) = 1 …(1)
On comparing with \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1
we have a² = 100; b² = 36
We know that, b² = a² (1 – e²)
⇒ 36 = 100 (1 – e²) ⇒ 1 – e² = \(\frac{36}{100}\) = \(\frac{9}{25}\)
⇒ e² = 1 – \(\frac{9}{25}\) = \(\frac{16}{25}\)
⇒ e = \(\frac{4}{5}\) (∵ e > 0)
required lengths of focal radii drawn through the point \(\left(8, \frac{18}{5}\right)\) = a ± ex₁
= 10 ± \(\frac{4}{5}\) × 8 = 10 ± \(\frac{32}{5}\) = \(\frac{82}{5}\), \(\frac{18}{5}\)
foci are given by (± ae, 0) i.e. \(\left( \pm 10 \times \frac{4}{5}, 0\right)\) i.e. (± 8, 0)
The eqns. of focal radii joining the points (± 8, 0) and \(\left(8, \frac{18}{5}\right)\) be given by

Question 21.
The focal distance of an end of the minor axis of the ellipse is k and the distance between the foci is 2h. Find the lengths of the semi-axes.
Solution:
Given, the focal distance of an end of minor axis be k.
The ends of minor axes be (0, ± b).
i.e. given a – ex₁ = k ⇒ a – e × 0 = k
⇒ a = k
distance between foci = 2h ⇒ 2ae = 2h
⇒ ae – h ⇒ e = \(\frac{h}{k}\)
We know that b² = a² (1 – e²)
⇒ b² = k² \(\left[1-\frac{h^2}{k^2}\right]\)
⇒ b² = k² – h²
⇒ b = \(\sqrt{k^2-h^2}\)

Question 22.
Find the eccentricity of the ellipse whose latus rectum is 4 and distance of the vertex from the nearest focus is 1.5 cm.
Solution:
Let e be the eccentricity of an ellipse
According to given condition, \(\frac{2 b^2}{a}\) = 4 ⇒ b² = 2a …(1)
Also, a – ae = 1.5 …(2)
Also we know that b² = a² (1 – e²) ⇒ 2a = a² (1 – e²) …(2)
⇒ 2a = a² – \(\left(a-\frac{3}{2}\right)^2\)
⇒ 2a = a² – a² – \(\frac{9}{4}\) + 3a ⇒ a = \(\frac{9}{4}\)
∴ from (2); \(\frac{9}{4}\) (1 – e) = \(\frac{3}{2}\)
⇒ 1 – e = \(\frac{3}{2}\) × \(\frac{4}{9}\)
⇒ 1 – e = \(\frac{2}{3}\) ⇒ e = \(\frac{1}{3}\)
Thus the required eccentricity of an ellipse be \(\frac{1}{3}\).

Question 23.
The directrix of a conic section is the line 3x + 4y = 1 and the focus S is (- 2, 3). If the eccentricity e is \(\frac{1}{\sqrt{2}}\), find the equation to the conic section.
Solution:
Given focus of ellipse be F (- 2, 3) and eqn. of corresponding directrix is 3x + 4y – 1 = 0 and eccentricity e = \(\frac{1}{\sqrt{2}}\).
Let P (x, y) be any point on ellipse and PM be the ⊥ drawn from P on given directrix. Then by definition of ellipse, we have | PF | = e | PM |
⇒ \(\sqrt{(x+2)^2+(y-3)^2}\) = \(\frac{1}{\sqrt{2}}\) \(\frac{|3 x+4 y-1|}{\sqrt{3^2+4^2}}\)
On squaring both sides ; we have
(x + 2)² + (y – 3)² = \(\frac{1}{2}\) × \(\frac{1}{25}\)(3x + 4y – 1)²
⇒ 50 [(x + 2)² + (y – 3)²] = (3x + 4y – 1)²
⇒ 50 (x² + y² + 4x – 6y + 13) = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 41x² + 34x² – 24xy + 206x – 292y + 649 = 0 which is the required eqn. of ellipse.

Question 24.
Find the equation to the conic section whose focus is (1, – 1), eccentricity is \(\left(\frac{1}{2}\right)\) and the directrix is the line x – y = 3. Is the conic section an ellipse ?
Solution:
Given focus of ellipse be F (1, – 1) and eqn. of directrix be x – y – 3 = 0 and e = \(\frac{1}{2}\)
Let P (x, y) be any point on ellipse. Then by def. of ellipse, we have | PF | = e | PM | where PM be the ⊥ drawn from P on given directrix.
⇒ \(\sqrt{(x-1)^2+(y+1)^2}\) = \(\frac{1}{2}\) \(\frac{|x-y-3|}{\sqrt{1^2+(-1)^2}}\)
On squaring both sides, we have
8 [(x – 1)² + (y + 1)²] = (x – y – 3)²
⇒ 8(x² – 2x + y² + 2y + 2) = x² + y² + 9 – 2xy + 6y – 6x
⇒ 7x² + 7y² + 2xy – 10x + 10y + 7 = 0 which is the required eqn. of ellipse.

Question 25.
Find the equation of the ellipse whose foci are (- 1, 5) and (5, 5) and whose major axis is 0.
Solution:
Given foci of an ellipse are S (- 1, 5) and S’ (5, 5).
Since ordinates of both foci are same.
Thus, eqn. of axes of an ellipse be parallel to x-axis and let C (α, β) be the coordinates of centre of an ellipse i.e. C (α, β) be the mid-point of SS’.
Thus, α = \(\frac{-1+5}{2}\) and β = \(\frac{5+5}{2}\)
i.e. a = 2 and p = 5
Let the required eqn. of ellipse be
\(\frac{(x-2)^2}{a^2}\) + \(\frac{(y-5)^2}{b^2}\) = 1 …(1)
given 2a = 10 ⇒ a = 5
Also, 2ae = 6 ⇒ ae = 3 ⇒ e = \(\frac{3}{5}\) < 1
We know that b² = a² (1 – e²)
⇒ b² = 25\(\left(1-\frac{9}{25}\right)\) = 16
Thus eqn. (1) reduces to ;
\(\frac{(x-2)^2}{25}\) + \(\frac{(y-5)^2}{16}\) = 1
which is the required eqn. of an ellipse.

Question 26.
Find the ellipse if its foci are (± 2, 0) and the length of the latus rectum is \(\frac{10}{3}\).
Solution:
Given foci of ellipse are (± 2, 0) and both foci lies on x-axis. So x-axis being the major axes of an ellipse.
Let the eqn. of ellipse be taken as
\(\frac{(x-0)^2}{a^2}\) + \(\frac{(y-0)^2}{b^2}\) = 1 …(1)
Clearly ae = 2 …(2)
and \(\frac{2 b^2}{a}\) = \(\frac{10}{3}\) ⇒ b² = \(\frac{5}{3}\)a …(3)
We know that b² = a² (1 – e²)
⇒ \(\frac{5a}{3}\) = a² – 4 [using eqn. (2) and (3)]
⇒ 3a² – 5a – 12 = 0
⇒ (a – 3) (3a + 4) = 0
⇒ a = 3 [∵ a > 0]
∴ from (2); e = \(\frac{2}{3}\) < 1
Thus b² = \(\frac{5}{3}\) × 3 = 5 [using (3)]
Hence eqn. (1) reduces to ; \(\frac{x^2}{9}\) + \(\frac{y^2}{5}\) = 1
which is the required eqn. of an ellipse.

Question 27.
Find the eccentricity of the ellipse of minor axis 2b, if the line segment joining the foci subtends an angle 2a at the upper vertex. Also, find the equation of the ellipse.
Solution:
Let e be the eccentricity of an ellipse.
In △BOS, we have OS = b tan α
∴ SS’ = 20S ⇒ 2ae – 2b tan α
⇒ ae = b tan α …(1)

We know that b² = a² (1 – e²)
⇒ b² = a² – a²e² = a² – b² tan² α
⇒ b² (1 + tan² α) = a²
a = b sec α …(2)
From eqn. (1) and eqn. (2); we have
e = \(\frac{b \tan \alpha}{b \sec \alpha}\) = sin α
Thus required eqn. of ellipse becomes ;
\(\frac{x^2}{b^2 \sec ^2 \alpha}\) + \(\frac{y^2}{b^2}\) = 1
⇒ x² cos² α + y² = b²
which is the required eqn.

Students can cross-reference their work with S Chand ISC Maths Class 11 Solutions Chapter 23 Parabola Chapter Test to ensure accuracy.

S Chand Class 11 ICSE Maths Solutions Chapter 23 Parabola Chapter Test

Question 1.
The equation of the directrix of the parabola is 3x + 2y + 1 = 0. The focus is (2, 1). Find the equation of the parabola.
Solution:
Given focus of parabola be F (2,1) and eqn. of directrix be 3x + 2y + 1 = 0.
Let P (x, y) be any point in the plane of cLiiectrix and focus.Let PM be the ⊥ drawn from P on given directrix.

On squaring both sides, we have
(x – 2)² + (y – 1)² = \(\frac{(3 x+2 y+1)^2}{13}\)
⇒ 13 [(x- 2)² + (y – 1)²] = (3x + 2y + 1)²
⇒ 13 (x² – 4x + y² – 2y + 5]
= 9x² + 4y² + 1 + 12xy + 4y + 6x
⇒ 4x² + 9y² – 12xy – 58x – 30y + 64 = 0
which is the required of parabola.

Question 2.
The points (0, 4) and (0, 2) are the vertex and focus of a parabola. Find the equation of the parabola.
Solution:
Given (0, 4) be the vertex and (0, 2) be focus of the parabola.
Since x-coordinate of both points be same. Hence axis of the parabola is y-axis. Further vertex (0, 4) lies above the focus (0, 2).

Thus the parabola be a downward parabola, length of latus rectum = 4a = 4 | VF | =4 × 2 = 8
Thus the required eqn. of parabola be
(x – 0)² = – 4a(y – 4)
⇒ x² = – 8 (y – 4)
⇒ x² + 8y – 32 = 0
which is required eqn. of parabola.

Question 3.
Find the equation of the parabola with latus rectum joining points (4, 6) and (4,-2).
Solution:
Given the ends of latus rectum are L (4, 6) and L’ (4, – 2).
Since the x-coordinates of both points L and L’ are equal.
∴ Latus-rectum is || to y-axis and hence axis of the parabola is parallel to x-axis.
[since axis ⊥ to latus rectum] Thus, the eqns. of two possible parabolas taken as, (y – k)² = ± 4a (x – h) …(1)
where (h, k) be the vertex of the parabolas.

length of latus-rectum = |LL’|
= \(\sqrt{(4-4)^2+(6+2)^2}\) = 8 = 4 a
∴ eqn. (1) reduces to,
(y – k)² = ± 8(x – h) …(2)
Since the point L (4, 6) lies on eqn. (2); we have
(6 – k)² = 8 (4 -h) …(3)
(6 – k)2 = – 8 (4 – h) …(4)
Also, the point L’ (4, – 2) lies on eqn. (2); we have

On dividing (3) and (5); we have
\(\frac{(6-k)^2}{(2+k)^2}\) = 1 ⇒ (6 – k)² = (2 + k)²
⇒ 36 + k² – 12k = k² + 4k + 4
⇒ 16k = 32 ⇒ k = 2
∴ from (5) ; (- 2 – 2)² = 8 (4 -h)
⇒ 16 = 8 (4 – h) ⇒ 2 = 4 – h ⇒ h = 2
Hence vertex of parabola be A (2, 2).
Thus eqn. of parabola be
(y – 2)² = 8(x – 2)
⇒ y² – 4y – 8x + 20 = 0
On dividing eqn. (4) and (6); we have
k = 2 ∴ from (4); 16 = -8(4 – h)
⇒ – 2 = 4 – h ⇒ h = 6
Thus vertex of required parabola be A’ (6, 2)
∴ eqn. of required parabola be given by (y – 2)² = – 8(x – 6)
⇒ y² – 4y + 8x – 44 = 0

Question 4.
Find the equation of the parabola whose focus is (- 1, – 2) and the equation of the directrix is given 4x – 3y + 2 = 0. Also find the equation of the axis.
Solution:
Given focus of parabola be F (- 1, – 2) and eqn. of directrix be

Let P (x, y) be any point in the plane of directrix and focus. Let PM be the 1 drawn from P (x, y) on given directrix. Then P (x, y) lies on parabola iff PF = PM

⇒ 25 [x² + 2x + y² + 4y + 5] = 16x² + 9y² + 4 – 24xy – 12y + 16x
⇒ 9x² + 16y² + 24xy + 34x + 112y + 121 = 0
which is the required eqn. of parabola. Since axis be the line ⊥ to directrix and passing through the focus of parabola, eqn. of line ⊥ to eqn. (1) be
3x + 4y + k = 0 …(2)
and eqn. (2) passes through the focus F (- 1, – 2)
∴ – 3 – 8 + k = 0
⇒ k = 11
∴ from (2); 3x + 4y + 11 = 0 be the required eqn. of axis of parabola.

Question 5.
Find the equation of the parabola if its vertex is at (0, 0), passes through (5, 2) and is symmetric w.r.t. y-axis.
Solution:
We know that a curve F (x, y) = 0 be symmetric w.r.t. y-axis if
F (-x, y) = F (x, y)
Let the eqn. of parabola having vertex (0, 0) and symmetric w.r.t. y be (x – 0)² = 4a(y – 0)
i.e. x² = 4ay …(1)
it passes through the point (5, 2); we have
25 = 8a ⇒ a = \(\frac { 25 }{ 8 }\)
Thus eqn. (1) reduces to ;
x² = 4 × \(\frac { 25 }{ 8 }\)y = \(\frac { 25 }{ 2 }\)y
which is the required eqn. of parabola.

Question 6.
The parabola y² = 4ax passes through the point (2, – 6). Find the length of its latus rectum.
Solution:
Given eqn. of parabola be
y² = 4ax …(1)
Now eqn. (1) passes through the point (2, – 6).
∴ 36 = 4a × 2 ⇒ a = \(\frac { 36 }{ 8 }\) = \(\frac { 9 }{ 2}\)
∴ length of latus rectum = 4a = 4 × \(\frac { 9 }{ 2}\) = 18.

Question 7.
Find the coordinates of the vertex and the focus of the parabola y² = 4(x + y).
Solution:
Equation of given parabola be
y² = 4 (x + y)
⇒ y² – 4y = 4x
⇒ y² – 4y + 4 = 4x + 4
⇒ (y – 2)² = 4 (x + 1) …(1)
shifting the origin to point (- 1, 2),
putting x + 1 =X and y – 2 = Y in eqn. (1); we have
Y² = 4X …(2)
Here 4a – 4
⇒ a = 1
which represents a right handed parabola with axis x-axis.
Vertex of parabola (2) be given by
X = 0; Y = 0
x + 1 = 0 and y – 2 = 0
i.e. x = – 1 and y = 2
Thus (-1, 2) be the vertex of parabola (1). Further, focus of parabola (2) be given by
X = a and Y = 0
⇒ x + 1 = 1 and y – 2 = 0
⇒ x = 0 and y = 2
Thus (0, 2) be the focus of parabola (1).

Question 8.
Find the focus, the equation of the directrix and the length of the latus rectum of the parabola y² + 12 = 4x+ 4y.
Solution:
Given eqn. of parabola be
y² – 4x – 4y + 12 = 0
⇒ y² – 4y = 4x – 12
⇒ y² – 4y + 4 = 4x – 8
⇒ (y – 2)² = 4 (x – 2) …(1)
Shifting the origin to point (2, 2) and putting
x – 2 = X and y – 2 = Y in eqn. (1);
we get Y² = 4X …(2)
which represents a right handed parabola.
On comparing with Y² = 4aX
∴ length of latus rectum = 4a = 4 units
Focus of eqn. (2) be given by
X = a, Y = 0
⇒ x – 2 = 1 and y – 2 = 0
i.e. x = 3 and y = 2
Thus (3, 2) are the coordinates of focus of eqn. (1).
eqn. of directrix of parabola (2) be given by X = – a ⇒ x – 2 = – 1 ⇒ x = 1
which is the eqn. of directrix.