Let the foci of the ellipse (\frac{x^2}{16}+\frac{y^2}{7}=1)... | Mathematics | SolveeIt | Solveeit

Today, August 18

Question

Let the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}$ coincide Then the length of the latus rectum of the hyperbola is:-

A

$\frac{32}{9}$

B

$\frac{18}{5}$

C

$\frac{27}{4}$

D

$\frac{27}{10}$

Answer

$\frac{27}{10}$

Explanation

Solution

16x2+7y2=1
⇒7=16(1−e2)⇒e=43
Foci of ellipse is (±ae,0)⇒(±3,0)
Now hyperbola be 144x2−αy2=251
25144x2−25αy2=1
Now a=512,b2=25α
Let eccentricity of hyperbola be e ae =3 (Given)
⇒512e=3⇒e=45
b2=a2(e2−1)
25α=25144(1625−1)⇒α=81
Hyperbola is 25144x2−2581y2=1
Now length of LR=a2b2=1027