Question

Let \[ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b \quad \text{and} \quad H: \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1. \] Let the distance between the foci of \( E \) and the foci of \( H \) be \( 2\sqrt{3} \). If \( a - A = 2 \), and the ratio of the eccentricities of \( E \) and \( H \) is \( \frac{1}{3} \), then the sum of the lengths of their latus rectums is equal to:

• A. 9
• B. 10
• C. 8
• D. 7

Hint:

For problems involving the foci and latus rectums of ellipses and hyperbolas: - Use the formulas for the eccentricity and latus rectum for both curves. - Apply given relationships, such as the distance between the foci or the ratio of eccentricities, to set up equations and solve for the unknowns.

Answer 1

The Correct Option is A

For the ellipse \( E \), the eccentricity \( e_E = \sqrt{1 - \frac{b^2}{a^2}} \) and the length of the latus rectum is \( \frac{2b^2}{a} \). 

For the hyperbola \( H \), the eccentricity \( e_H = \sqrt{1 + \frac{B^2}{A^2}} \) and the length of the latus rectum is \( \frac{2B^2}{A} \). 

Given that the ratio of the eccentricities of \( E \) and \( H \) is \( \frac{1}{3} \), and using the condition \( a - A = 2 \), we can set up equations to solve for the required lengths of the latus rectums. The sum of these lengths is \( 9 \).

 Thus, the answer is 9.