Question

If a variable line has its intercepts on the co-ordinate axes $e,e^{\prime }$, where $\frac{e}{2},\frac{e^{\prime }}{2}$ are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle $x^2+y^2=r^2$, where $r=$

Answer 1

Since $\frac{e}{2}$ and $\frac{e^{{}^{\prime }}}{2}$ are the eccentricities of a hyperbola and its conjugate, we have
$\frac{4}{e^2}+\frac{4}{e^{{}^{\prime }2}}=1$
or $4=\frac{e^2{e}^{{}^{\prime }2}}{e^{{{}^{\prime }}^2}+e^{{{}^{\prime }}^2}}$
The line is passing through the points $(e,0)$ and $(0,e^{{}^{\prime }})$. Therefore,
$e^{{}^{\prime }}x+ey-e{e}^{{}^{\prime }}=0$
It is tangent to the circle
$x^2+y^2=r^2$
$\therefore \frac{e{e}^{{}^{\prime }}}{\sqrt{e^2+e^{{}^{\prime }2}}}=r$
or $r^2=\frac{e^2{e}^{{}^{\prime }2}}{e^2+e^{{}^{\prime }2}}=4$
or $r=2$