Let the ellipse E:x^2144+y^2169=1 and the hyperbola H:x^216-y^2^2=-1 have the same foci.

Question

Let the ellipse $E:\dfrac{x^2}{144}+\dfrac{y^2}{169}=1$ and the hyperbola $H:\dfrac{x^2}{16}-\dfrac{y^2}{\lambda^2}=-1$ 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:

Accepted Answer

For $E$: $a^2=144,\ b^2=169$ (major axis along $y$ since $169>144$), so $c^2=169-144=25$, foci $(0,\pm5)$.

Rewrite $H$: $\dfrac{y^2}{\lambda^2}-\dfrac{x^2}{16}=1$, a hyperbola with transverse semi-axis $B=\lambda$ (along $y$) and $A^2=16$. Same foci give $c^2=A^2+B^2=25 \Rightarrow \lambda^2=9\Rightarrow\lambda=3$.

$$e=\frac{c}{B}=\frac{5}{3},\qquad L=\frac{2A^2}{B}=\frac{2\times16}{3}=\frac{32}{3}.$$

$$24(e+L)=24\left(\frac53+\frac{32}{3}\right)=24\times\frac{37}{3}=296.$$

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