Let E:x^2a^2+y^2b^2=1 (ab) and H:x^2A^2-y^2B^2=1.

Question

Let $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b)$ and $H:\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$. Let the distance between the foci of $E$ and the (distance between the) foci of $H$ each be $2\sqrt3$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $1:3$, then the sum of the lengths of their latus recta is equal to:

Accepted Answer

$$2\sqrt{a^2-b^2}=2\sqrt3 \Rightarrow a^2-b^2=3,\qquad 2\sqrt{A^2+B^2}=2\sqrt3 \Rightarrow A^2+B^2=3.$$

$$e_E=\frac{\sqrt3}{a},\quad e_H=\frac{\sqrt3}{A},\qquad \frac{e_E}{e_H}=\frac{A}{a}=\frac13 \Rightarrow a=3A.$$

With $a-A=2$: $2A=2\Rightarrow A=1,\ a=3$. Then $b^2=9-3=6$, $B^2=3-1=2$.

$$L_E=\frac{2b^2}{a}=\frac{12}{3}=4,\qquad L_H=\frac{2B^2}{A}=\frac{4}{1}=4,\qquad L_E+L_H=8.$$

Solutions · 1