Question

If we rotate the axes of the rectangular hyperbola $x^2-y^2=a^2$ through an angle $\pi / 4$ in the clockwise direction then the equation $x^2-y^2=a^2$ reduces to $x y=\frac{a^2}{2}=\left(\frac{a}{\sqrt{2}}\right)^2=c^2$ (say). Since $x=c t, y=\frac{c}{t}$ satisfies $x y=c^2$
$\therefore (x, y)=\left(c t, \frac{c}{t}\right)(t \neq c)$ is called a 't' point on the rectangular hyperbola.
If $e_1$ and $e_2$ are the eccentricities of the hyperbolas $x y=9$ and $x^2-y^2=25$, then $\left(e_1\right.$, e $\left.e_2\right)$ lie on a circle $C _1$ with centre origin then the (radius) ${ }^2$ of the director circle of $C _1$ is -

• A. 2
• B. 4
• C. 8
• D. 16

Answer 1

Answer: (C)

$e_1=\sqrt{2}, e_2=\sqrt{2}$
$\Rightarrow (\sqrt{2}, \sqrt{2})$ is the point on the circle.
$\Rightarrow $ radius of $C_1=2$.
$\Rightarrow $ radius of director circle of $C_1=2 \sqrt{2}$.
$\therefore (\text { radius) })^2=8$