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

Correct Answer: 51

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

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

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

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

\( \implies \sqrt{2} - \sqrt{2}e^2 = e \)

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

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

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

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

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