Question

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

Answer 1

$\text{foci} \equiv (\pm 5, 0); \quad \frac{2b^2}{a} = \sqrt{50}$

$ae = 5 \quad \text{and} \quad b^2$
$= \frac{5\sqrt{2}a}{2} b^2 = a^2(1 - e^2)$
$= \frac{5\sqrt{2}a}{2}$

$\implies a(1 - e^2)$
$= \frac{5\sqrt{2}}{2}$

$\implies \frac{5}{e(1 - e^2)}$
$= \frac{5\sqrt{2}}{2}$

$\implies \sqrt{2} - \sqrt{2}e^2 = e$

$\implies \sqrt{2}e^2 + e - \sqrt{2} = 0$

$\implies \sqrt{2}(e + \sqrt{2}) - 1(1 + \sqrt{2}) = 0$

$\implies (e + \sqrt{2})(\sqrt{2}e - 1) = 0$

$\therefore e = \sqrt{2}, \quad e = \frac{1}{\sqrt{2}}$

For the hyperbola: $\frac{x^2}{b^2} - \frac{y^2}{a^2b^2} = 1$ $a = 5\sqrt{2}, \quad b = 5$ $a^2b^2 = b^2(e_1^2 - 1) \implies e_1^2 = 51$