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

Equation of the ellipse is
$\frac{x^2}{25}+\frac{y^2}{9}=1(a>b)$
Foci (±4,0) 

Let S′(h,k) be the image.
Mid point of SS′ is M=$(\frac{h\pm 4}{2},\frac{k}{2})$
SS′ cuts tangent line at point M 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 $x^2+y^2=25$
∴M lies on auxiliary circle.
So,
⇒$$\begin{gathered} \frac{{(h\pm 4)}^2}{4}+\frac{k^2}{4}=25 \\ \Rightarrow {(h\pm 4)}^2+k^2=100 \end{gathered}$$
∴ Required locus equation is ${(x\pm 4)}^{2_{ }}+y^2=100$