The locus of the feet of the perpendiculars from centre of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ to any tangent at

let the ellipse be $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

let P(h, k) be the foot of perpendicular to a tangent $\mathrm{y}=\mathrm{mx}\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$ from centre.

$$\begin{gathered} \Rightarrow \frac{\mathrm{k}}{h} x m=-1 \\ \Rightarrow \mathrm{m}=\frac{-\mathrm{h}}{\mathrm{k}} \end{gathered}$$

since P(h, k) lies on tangent then 

$\begin{array}{rcl}\mathrm{k}& =& \mathrm{mh}+\sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}\\ \mathrm{k}& =& \frac{-{\mathrm{h}}^2}{\mathrm{k}}+\sqrt{{\mathrm{a}}^2\left(\frac{{\mathrm{h}}^2}{{\mathrm{k}}^2}\right)+{\mathrm{b}}^2}\\ & \Rightarrow & {\left(\mathrm{k}+\frac{{\mathrm{h}}^2}{\mathrm{k}}\right)}^2=\frac{{\mathrm{a}}^2{\mathrm{h}}^2}{{\mathrm{k}}^2}+{\mathrm{b}}^2\\ & \Rightarrow & {\left(\mathrm{k}+\frac{{\mathrm{h}}^2}{\mathrm{k}}\right)}^2=\frac{{\mathrm{a}}^2{\mathrm{h}}^2}{{\mathrm{k}}^2}+{\mathrm{b}}^2\\ & \Rightarrow & \frac{{({\mathrm{k}}^2+{\mathrm{h}}^2)}^2}{{\mathrm{k}}^2}=\frac{{\mathrm{a}}^2{\mathrm{h}}^2+{\mathrm{b}}^2{\mathrm{k}}^2}{{\mathrm{k}}^2}\end{array}$

$\therefore$ locus of P(h, k) is ${({\mathrm{x}}^2+{\mathrm{y}}^2)}^2={\mathrm{a}}^2{\mathrm{x}}^2+{\mathrm{b}}^2{\mathrm{y}}^2$