Question

If $l$ is the length of the intercept made by a common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$ on the coordinate axes, then $\frac{3l^2}{28}$ is equal to

Answer 1

The equation of any tangent to the ellipse is $y=mx+\sqrt{25m^2+4}$, which also touches the circle if the distance of this line from center $(0,0)$ is equal to radius of the circle.
$\left|\frac{m\left(0\right)-0+\sqrt{25m^2+4}}{\sqrt{1+m^2}}\right|=4$
$\Rightarrow$ $25m^2+4=16(1+m^2)$

$\Rightarrow m^2=\frac{4}{3}$
$\Rightarrow$ $m=\frac{\pm 2}{\sqrt{3}}$ and the equation of the common tangent becomes
$y=\pm \frac{2}{\sqrt{3}}x+\sqrt{25\times \frac{4}{3}+4}$
$\Rightarrow$ $\sqrt{3}y\pm 2x=4\sqrt{7}$
This tangent intersects coordinate axes at $(\frac{4\sqrt{7}}{2},0)$ and $(0,\frac{4\sqrt{7}}{\sqrt{3}})$
Distance between these intersection points is $l$.
$l^2={\left(\frac{4\sqrt{7}}{2}\right)}^2+{\left(\frac{4\sqrt{7}}{\sqrt{3}}\right)}^2$
$\Rightarrow$ $3l^2=196\Rightarrow \frac{3l^2}{28}=7$
Answer: 7