The equation of a hyperbola whose asymptotes are 3x ± 5y = 0 and vertices are ( ± 5, 0) is

Question

The equation of a hyperbola whose asymptotes are 3x ± 5y = 0 and vertices are ( ± 5, 0) is (A) 3x2-5y2=25 (B) 5x2 - 3y2 = 225 (C) 25x2-9y2=225 (D) 9x2 -25y2 =225

Tardigrade Admin · Accepted Answer

The given equation of asymptotes of hyperbola is

3 x \pm 5 y = 0

...(i)
and vertices

(\pm 5, 0)

Now, equation of hyperbola is

(3 x + 5 y) (3 x - 5 y) = λ

9 x^{2} - 25 y^{2} = λ

...(ii)
Since,

(\pm 5, 0)

is the vertices of the hyperbola, then

9 (5)^{2} - 0 = λ \Rightarrow λ = 225

From Eq. (ii),

9 x^{2} - 25 y^{2} = 225

Q. The equation of a hyperbola whose asymptotes are $3 x \pm 5 y = 0$ and vertices are $(\pm 5, 0)$ is

$3 x^{2} - 5 y^{2} = 25$

$5 x^{2} - 3 y^{2} = 225$

$25 x^{2} - 9 y^{2} = 225$

$9 x^{2} - 25 y^{2} = 225$

Q. The equation of a hyperbola whose asymptotes are 3x±5y=0 and vertices are (±5,0) is

3x2−5y2=25

5x2−3y2=225

25x2−9y2=225

9x2−25y2=225

The given equation of asymptotes of hyperbola is 3x±5y=0...(i) and vertices (±5,0) Now, equation of hyperbola is (3x+5y)(3x−5y)=λ 9x2−25y2=λ...(ii) Since, (±5,0) is the vertices of the hyperbola, then 9(5)2−0=λ⇒λ=225 From Eq. (ii), 9x2−25y2=225

Q. The equation of a hyperbola whose asymptotes are $3 x \pm 5 y = 0$ and vertices are $(\pm 5, 0)$ is

$3 x^{2} - 5 y^{2} = 25$

$5 x^{2} - 3 y^{2} = 225$

$25 x^{2} - 9 y^{2} = 225$

$9 x^{2} - 25 y^{2} = 225$