OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Ex 25(b)

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