Question

Let the hyperbola
$H:\frac{x^2}{a^2}−y^2=1$
and the ellipse
$E:3x^2+4y^2=12$
be such that the length of latus rectum of H is equal to the length of latus rectum of E. If eH and eE are the eccentricities of H and E respectively, then the value of
$12 (e^{2}_H+e^{2}_E)$ is equal to _____ .

Answer 1

The correct answer is 42
$∴ H:\frac{x^2}{a^2}−\frac{y^2}{1}=1$
∴ Length of latus rectum
$=\frac{2}{a}$
$E:3x^2+4y^2=12$
Length of latus rectum
$= \frac{6}{2} = 3$
$\because \frac{2}{a} = 3 ⇒ a = \frac{2}{3}$
$12 (e^{2}_H+e^{2}_E)$
$= 12(1+\frac{9}{4})+(1-\frac{3}{4})$
$= 42$