Question

A straight line $PQ$ touches the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and the circle $x^2+y^2=r^2(3<r<4).$ $RS$ is chord of the circle which is parallel to $PQ$ and passes through any focus of ellipse, then the length of $RS$ is equal to unit.

Answer 1

$y=mx\pm \sqrt{16m^2+9}$ is tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1.$

If this is also tangent to the circle $x^2+y^2=r^2,$ then
$r^2=\frac{16m^2+9}{1+m^2}m=\sqrt{\frac{r^2-9}{16-r^2}}(\because 3<r<4)\dots \left(i\right)$

Equation of line $RS$ be
$y-0=\sqrt{\frac{r^2-9}{16-r^2}}(x+\sqrt{7})$

Length of chord drawn from centre $C$ of circle to chord
$=2\sqrt{r^2-d^2}$
Where, $d=$ perpendicular distance from centre to the chord
$=\left|\frac{\sqrt{7}\sqrt{\frac{r^2-9}{16-r^2}}}{\sqrt{1^2+\frac{r^2-9}{16-r^2}}}\right|$
$=\sqrt{r^2-9}$

Hence, required length of chord $=2\sqrt{r^2-(\sqrt{r^2-9}{)}^2}$
$=6$ units.