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 1

Answer: (D)

$\frac{x^2}{16}+\frac{y^2}{7}=1$
$\Rightarrow 7=16\left(1-e^2\right)\Rightarrow e=\frac{3}{4}$
Foci of ellipse is $(\pm ae,0)\Rightarrow (\pm 3,0)$
Now hyperbola be $\frac{x^2}{144}-\frac{y^2}{\alpha }=\frac{1}{25}$
$\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{\alpha }{25}}=1$
Now $a=\frac{12}{5},b^2=\frac{\alpha }{25}$
Let eccentricity of hyperbola be e ae $=3$ (Given)
$\Rightarrow \frac{12}{5}e=3\Rightarrow e=\frac{5}{4}$
$b^2=a^2\left(e^2-1\right)$
$\frac{\alpha }{25}=\frac{144}{25}\left(\frac{25}{16}-1\right)\Rightarrow \alpha =81$
Hyperbola is $\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1$
Now length of $LR=\frac{2b^2}{a}=\frac{27}{10}$