Let the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and the hyperbola $\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$,

Question

Let the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and the hyperbola $\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$ be such that their respective end points of latus rectum coincide, then sum of squares of their eccentricities is equal to

(A) $\dfrac{7}{4}$

(B) $2$

(C) $\dfrac{11}{4}$

(D) None

My Approach:

Let eccentricity of ellipse and hyperbola be $e$ and $E$ respectively.

It is given that $ae=AE$ and $\dfrac{b^2}{a}=\dfrac{B^2}{A}.......(1)$

Hence $b^2=a^2(1-e^2)$ and $B^2=A^2(E^2-1)$

$\implies e^2=1-\dfrac{b^2}{a^2}$ and $E^2=\dfrac{B^2}{A^2}-1$

$\implies e^2+E^2=\dfrac{B^2}{A^2}-\dfrac{b^2}{a^2}$

$\implies e^2+E^2=\dfrac{b^2}{A^2}\bigg(\dfrac{B^2}{b^2}-\dfrac{A^2}{a^2}\bigg)$

Using equation $(1)$ we get $e^2+E^2=\dfrac{b^2}{A^2}\bigg(\dfrac{A}{a}-\dfrac{A^2}{a^2}\bigg)$

How to proceed from here?

Answer 1

Solving as you did, $$e=\sqrt{(A/a)}$$ $$E=\sqrt{(a/A)}$$ Therefore the value wanted in the question is: $$\frac{a}{A}+\frac{A}{a}$$ However, without any restrictions given on $a$ or $A$ or any relation between them, we cannot tell any definite value for the squares of eccentricities. So the best possible option to your question is (d)None.

Also as you can see here, changing $a$ or $A$ changes the sum of squares of the eccentricities.