Let $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and $H: \frac{x^2}{A^2} – \frac{y^2}{B^2} = 1$. 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:
- (1) 10
- (2) 7
- (3) 8
- (4) 9
Solution:
Foci of E: $2ae = 2\sqrt{3} \Rightarrow ae = \sqrt{3}$.
Foci of H: $2Ae’ = 2\sqrt{3} \Rightarrow Ae’ = \sqrt{3}$.
So $ae = Ae’$.
Ratio of eccentricities $\frac{e}{e’} = \frac{1}{3} \Rightarrow e’ = 3e$.
Substitute: $a e = A (3e) \Rightarrow a = 3A$.
Given $a – A = 2 \Rightarrow 3A – A = 2 \Rightarrow 2A = 2 \Rightarrow A = 1$.
Then $a = 3$.
Find e: $ae = \sqrt{3} \Rightarrow 3e = \sqrt{3} \Rightarrow e = \frac{1}{\sqrt{3}}$.
Find e’: $e’ = 3e = \sqrt{3}$.
For Ellipse E: $b^2 = a^2(1-e^2) = 9(1 – 1/3) = 9(2/3) = 6$.
Latus Rectum of E $= \frac{2b^2}{a} = \frac{2(6)}{3} = 4$.
For Hyperbola H: $B^2 = A^2(e’^2 – 1) = 1(3 – 1) = 2$.
Latus Rectum of H $= \frac{2B^2}{A} = \frac{2(2)}{1} = 4$.
Sum of Latus Rectums $= 4 + 4 = 8$.
Ans. (3)