Let the ellipse E: x²/144+y²/169=1 and the hyperbola H: x²/16-y²/… | JEE Main 2026 Mathematics PYQ

MCQMediumJEE Main 2026 · 28 January, Shift 2Conic Sections (Parabola, Ellipse, Hyperbola)

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:

A
$67$
B
$296$
C
$148$
D
$126$

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