OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test

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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.