The locus of the foot of perpendicular from the centre on any tangent to be ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$is 

$$\begin{gathered} Let P (h,k) be the foot of perpendicular to a \tan gent y=mx+\sqrt{a^2{m}^2+b^2} from center \\ \therefore \frac{k}{h}.m=-1\Rightarrow m=-\frac{h}{k} \\ \because P(h,k) lies on \tan gent \\ \therefore k=mh+\sqrt{a^2{m}^2+b^2 } \\ sub m value , we get {(k+\frac{h^2}{k})}^2=\frac{a^2{h}^2+b^2}{k^2} \\ \Rightarrow locus is {(x^2+y^2)}^2=a^2{x}^2+b^2{y}^2 \end{gathered}$$