Ellipse
Ellipse-Hyperbola Common Foci — Eccentricity and Latus Rectum
nta_pyq_2026_jan
Grade 11
Question:
Let the ellipse $E:\dfrac{x^2}{144}+\dfrac{y^2}{169}=1$ and the hyperbola $H:\dfrac{y^2}{z^2}-\dfrac{x^2}{\lambda^2}=-1$ 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:
Step-by-Step Solution
Key Concept: Ellipse $E$: $a^2=169$, $b^2=144$, $c_E=\sqrt{169-144}=5$. Foci of $E$ at $(0,\pm5)$. Hyperbola $H$: $\dfrac{y^2}{z^2}-\dfrac{x^2}{\lambda^2}=1$ with $c_H=5$. So $z^2+\lambda^2=25$.
$e=\tfrac{5}{3}$, $L=\tfrac{32}{3}$. $24(e+L)=296$.
Correct Answer: 4