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

The Correct Option is C

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: \( \frac{10\sqrt{5}}{3} \)