Question

The slope of common tangent to the curves $x^2+y^2=16$ and $\frac{x^2}{25}+\frac{y^2}{4}=1$ in the $1^{\text{st}}$ quadrant is $m$, then $3m^2=$

Answer 1

Any tangent to ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$ with slope $m$ is :
$y=mx\pm \sqrt{25m^2+4}\cdots \left(1\right)$
If this is also a tangent to the circle $x^2+y^2=16$, then
Distance of tangent from centre of cricle $=4$
$\Rightarrow \left|\frac{0-m\times 0\pm \sqrt{25m^2+4}}{\sqrt{1+m^2}}\right|=4$
$\Rightarrow m=\pm \frac{2}{\sqrt{3}}$
Since common tangent is in the first quadrant, so angle will be obtuse
$m=-\frac{2}{\sqrt{3}}$
$\therefore 3m^2=4$