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. $2e_h^2-e_e^2=6$
• B. $e_e^2-4e_h^2=6$
• C. $4e_h^2-e_e^2=6$
• D. $e_h^2-2e_e^2=0$

Answer 1

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

Equation of director circles of ellipse and hyperbola are $x^2+y^2=a^2+b^2$ and $x^2+y^2=a^2-b^2$ respectively.
So,
$a^2+b^2=4r^2\cdots \left(1\right)$
$a^2-b^2=r^2\cdots \left(2\right)$
Solving above equations, we get $a^2=\frac{5r^2}{2},b^2=\frac{3r^2}{2}$
Now,
$e_e^2=1-\frac{b^2}{a^2}$
$\Rightarrow e_e^2=1-\frac{3r^2}{2}\times \frac{2}{5r^2}=1-\frac{3}{5}=\frac{2}{5};$
$e_h^2=1+\frac{b^2}{a^2}$
$\Rightarrow e_h^2=1+\frac{3}{5}=\frac{8}{5}$
So, from given options$4e_h^2-e_e^2=4\times \frac{8}{5}-\frac{2}{5}=\frac{30}{5}=6$