Question

If radii of director circles of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $2r$ and $r$ respectively and $e_e$ and $e_h$ be the eccentricities of the ellipse and the hyperbola respectively then

• A. $2{e_h}^2-{e_e}^2=6$
• B. ${e_e}^2-{4e_h}^2=6$
• C. ${4e_h}^2-{e_e}^2=6$
• D. none of these

Answer 1

The correct option is C ${4e_h}^2-{e_e}^2=6$

Equation of director circles of ellipse and hyperbola are respectively,
$x^2+y^2=a^2+b^2$ and $x^2+y^2=a^2-b^2$
$a^2+b^2=4r^2......\left(1\right)$
$a^2-b^2=r^2......\left(2\right)$
So, $2a^2=5r^2$
$a^2=\frac{5r^2}{2}$

$b^2=\frac{3r^2}{2}$
$\therefore {e_e}^2=1-\frac{b^2}{a^2}$

$\Rightarrow {e_e}^2=1-\frac{3r^2}{2}\times \frac{2}{5r^2}=1-\frac{3}{5}=\frac{2}{5}$...........$(\because {e_h}^2=1+\frac{b^2}{a^2}$)

$\Rightarrow {e_h}^2=1+\frac{3}{5}=\frac{8}{5}$

$\Rightarrow {4e_h}^2-{e_e}^2=4\times \frac{8}{5}-\frac{2}{5}=\frac{30}{5}=6$