Question

Let $e_1$ be the eccentricity of the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ and $e_2$ be the eccentricity of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b,$ which passes through the foci of the hyperbola. If $e_1 e_2 = 1$, then the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is:

• A. $4\sqrt{5}$
• B. $\frac{8\sqrt{5}}{3}$
• C. $\frac{10\sqrt{5}}{3}$
• D. $3\sqrt{5}$

Answer 1

Answer: (C)

$\frac{10\sqrt{5}}{3}$

Answer 2

Given:
$\frac{x^2}{16} + \frac{y^2}{9} = 1 \implies e_1 = \sqrt{1 - \frac{9}{16}} = \frac{5}{4}.$

For the ellipse:
$e_1 e_2 = 1 \implies e_2 = \frac{4}{5}.$

The ellipse passes through $(\pm 5, 0)$, so $a = 5$ and $b = 3$:
$\frac{x^2}{25} + \frac{y^2}{9} = 1.$

The length of the chord parallel to the $x$-axis and passing through $(0, 2)$ is given by:
$L = 2a \sqrt{1 - \frac{y^2}{b^2}} = 2 \times 5 \times \sqrt{1 - \frac{4}{9}} = 10 \sqrt{\frac{5}{9}} = \frac{10 \sqrt{5}}{3}.$

The Correct answer is: $\frac{10\sqrt{5}}{3}$