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

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.