Let the equation of the ellipse be $\mathrm{S}=\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}-1=0$

Let P(x₁, y₁) be the foot of the perpendicular drawn from either of the foci to a tangent.
The equation of the tangent to the ellipse $\mathrm{S}=0\text{ is }\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2} ...\left(1\right)$
The equation to the perpendicular from either foci ($\pm$ae,0) on this tangent is
$\mathrm{y}=-\frac{1}{\mathrm{m}}(\mathrm{x}\pm \mathrm{ae})\dots \left(2\right)$
As P is the point of intersection of (1) & (2)
we have
$\begin{array}{l}{\mathrm{y}}_1={\mathrm{mx}}_1\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2},{\mathrm{y}}_1=-\frac{1}{\mathrm{m}}\left({\mathrm{x}}_1\pm \mathrm{ae}\right)\\ \Rightarrow {\mathrm{y}}_1-{\mathrm{mx}}_1=\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2},{\mathrm{my}}_1+{\mathrm{x}}_1=\pm \mathrm{ae}\\ \Rightarrow {\left({\mathrm{y}}_1-{\mathrm{mx}}_1\right)}^2+{\left({\mathrm{my}}_1+{\mathrm{x}}_1\right)}^2={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2+{\mathrm{a}}^2{\mathrm{e}}^2\\ \Rightarrow {\mathrm{y}}_1^2+{\mathrm{m}}^2{\mathrm{x}}_1^2-2{\mathrm{x}}_1{\mathrm{y}}_1\mathrm{m}+{\mathrm{m}}^2{\mathrm{y}}_1^2+{\mathrm{x}}_1^2+2{\mathrm{x}}_1{\mathrm{y}}_1\mathrm{m}\\ ={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)+{\mathrm{a}}^2{\mathrm{e}}^2\end{array}$
$\begin{array}{l}\Rightarrow {\mathrm{x}}_1^2\left({\mathrm{m}}^2+1\right)+{\mathrm{y}}_1^2\left(1+{\mathrm{m}}^2\right)={\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{a}}^2\\ \Rightarrow \left({\mathrm{x}}_1^2+{\mathrm{y}}_1^2\right)\left({\mathrm{m}}^2+1\right)={\mathrm{a}}^2\left({\mathrm{m}}^2+1\right)\\ \Rightarrow {\mathrm{x}}_1^2+{\mathrm{y}}_1^2={\mathrm{a}}^2\end{array}$
$\therefore \mathrm{P}\text{ lies on }{\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$ which is a circle with centre at the origin, the centre of ellipse.