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. 8
• B. 10
• C. 9
• 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

The given equations are: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{(foci are } (ae, 0) \text{ and } (-ae, 0)) \] \[ \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1 \quad \text{(foci are } (Ae', 0) \text{ and } (-Ae', 0)) \] From the above, we have: \[ 2ae = 2\sqrt{3} \quad \Rightarrow \quad ae = \sqrt{3} \] Also: \[ 2Ae' = 2\sqrt{3} \quad \Rightarrow \quad Ae' = \sqrt{3} \] So, we get: \[ ae = Ae' \quad \Rightarrow \quad \frac{e}{e'} = \frac{A}{a} \] This gives: \[ \frac{1}{3} = \frac{A}{a} \quad \Rightarrow \quad a = 3A \] Now, using \( a - A = 2 \), we have: \[ a - A = 2 \quad \Rightarrow \quad a = 3 \quad \text{and} \quad A = 1 \] Substituting into the equation \( A = \sqrt{3} \), we get: \[ A = \sqrt{3}, \quad e = \frac{1}{\sqrt{3}}, \quad e' = \sqrt{3} \] Now, for the semi-major axis \( b^2 \), we have: \[ b^2 = a^2 (1 - e^2) \] \[ b^2 = 6 \] For the semi-major axis of the hyperbola \( B^2 \): \[ B^2 = A^2 \left( (e')^2 - 1 \right) \] \[ B^2 = 2 \] Finally, the sum of the lengths of the latus rectums for both the ellipse and the hyperbola is: \[ \text{Sum of LR} = \frac{2b^2}{a} + \frac{2B^2}{A} = 8 \]