Q. Let the ellipse $E$ :

$ \frac{x^{2}}{144}+\frac{y^{2}}{169}=1 $

and the hyperbola $H$ :

$ \frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1 $

have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of $H$, then the value of $24(e+L)$ is :

Correct Answer: 296

Explanation

For the ellipse

$\dfrac{x^{2}}{144}+\dfrac{y^{2}}{169}=1$

Here,

$a^{2}=169,\quad b^{2}=144$

So,

$c^{2}=a^{2}-b^{2}=169-144=25 \Rightarrow c=5$

Thus, the distance of each focus from the center is $5$.

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Now consider the hyperbola

$\dfrac{x^{2}}{16}-\dfrac{y^{2}}{\lambda^{2}}=-1$

Rewriting in standard form,

$\dfrac{y^{2}}{\lambda^{2}}-\dfrac{x^{2}}{16}=1$

So,

$a^{2}=\lambda^{2},\quad b^{2}=16$

For a hyperbola,

$c^{2}=a^{2}+b^{2}=\lambda^{2}+16$

Since both conics have the same foci,

$c=5 \Rightarrow \lambda^{2}+16=25 \Rightarrow \lambda^{2}=9$

Hence,

$a=3,\quad b=4$

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Eccentricity of the hyperbola:

$e=\dfrac{c}{a}=\dfrac{5}{3}$

Length of the latus rectum of the hyperbola:

$L=\dfrac{2b^{2}}{a}=\dfrac{2\times16}{3}=\dfrac{32}{3}$

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Now,

$e+L=\dfrac{5}{3}+\dfrac{32}{3}=\dfrac{37}{3}$

Therefore,

$24(e+L)=24\times\dfrac{37}{3}=8\times37=296$

Hence, the correct answer is

296

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