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Hyperbola

JEE Main 2026 (Online) 28th January Evening Shift

MCQ+4 / -12026

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 :

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