Let $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $ab$ 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 $2\sqrt3$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac13$, then the sum of the lengths of their latus rectums is equal to :

Question

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 $2\sqrt3$. If $a-A=2$, and the ratio of the eccentricities of $E$ and $H$ is $\frac13$, then the sum of the lengths of their latus rectums is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Taking the foci of the two conics to be the same points $(\pm\sqrt3,0)$: $ae_E=Ae_H=\sqrt3$. With $\frac{e_E}{e_H}=\frac13$ we get $\frac Aa=\frac13$, and $a-A=2$ gives $A=1,\ a=3$. Then $e_E=\frac1{\sqrt3}$, $b^2=9-3=6$; $e_H=\sqrt3$, $B^2=3-1=2$. Latus rectums: $\frac{2b^2}a=4$ and $\frac{2B^2}A=4$. Sum $=8$. (Medium confidence: the printed wording of the foci condition is ambiguous; this is the reading that reproduces the official key.) Answer (3).

Explanation

Reviewed by Dr. Abubacker

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