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 $xy=\frac{a^2}{2}={\left(\frac{a}{\sqrt{2}}\right)}^2=c^2$ (say). Since $x=ct,y=\frac{c}{t}$ satisfies $xy=c^2$
$\therefore (x,y)=\left(ct,\frac{c}{t}\right)(t\ne c)$ is called a 't' point on the rectangular hyperbola.
If $e_1$ and $e_2$ are the eccentricities of the hyperbolas $xy=9$ and $x^2-y^2=25$, then $(e_1$, e $e_2)$ 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 $C_1=2\sqrt{2}$.
$\therefore (\text{ radius) }{)}^2=8$