Question

If a variable line has its intercepts on the coordinates axes as $e,e^{\prime }$ where $\frac{e}{2},\frac{e^{\prime }}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively, then the line always touches the circle centred at $O$ whose radius $r$ is equal to

Answer 1

Since, $\frac{e}{2}$ and $\frac{e^{\prime }}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively.
$\therefore \frac{4}{e^2}+\frac{4}{(e^{\prime }{)}^2}=1$
$\Rightarrow \frac{1}{e^2}+\frac{1}{(e^{\prime }{)}^2}=\frac{1}{4}$
Let line equation having intercepts as $e,e^{\prime }$ be $\frac{x}{e}+\frac{y}{e^{\prime }}=1$
since it is the tangent to the circle $x^2+y^2=r^2,$ so perpendicular distance from origin to the tangent is the radius of the circle.
$\therefore \left|\frac{1}{\sqrt{\frac{1}{e^2}+\frac{1}{(e^{\prime }{)}^2}}}\right|=r$
$\Rightarrow r=2$ units