Let the foci and length of the latus rectum of an ellipse x²/a² +… | JEE Main 2024 Mathematics PYQ

MCQMediumJEE Main 2024 · 31 January, Shift 1Conic Sections (Parabola, Ellipse, Hyperbola)

Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , $a > b$ be $(\pm 5, 0)$ and $\sqrt{50}$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$ equals:

A
$51$
B
$48$
C
$50$
D
$45$

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