Question

The locus of the image of the focus of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ with respect to any of the tangents to the ellipse is

Answer 1

Equation of the ellipse is
$\frac{x^2}{25}+\frac{y^2}{9}=1$
Foci $(\pm 4,0)$
Let $S^{\prime }(h,k)$ be the image.
$\text{Mid point of}S{S}^{\prime }\equiv (\frac{h\pm 4}{2},\frac{k}{2})$
$S{S}^{\prime }$ cuts tangent line at point which lies on the auxiliary circle of the ellipse , since foot of perpendicular from foci upon any tangent lies on auxilary circle.
Auxiliary circle of the ellipse is $x^2+y^2=25$

So,
$\Rightarrow {\left(\frac{h\pm 4}{2}\right)}^2+\frac{k^2}{4}=25$
$\therefore$ Required locus equation is $(x\pm 4{)}^2+y^2=100$