Question

Find the locus of foot of perpendicular drawn from centre to any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Answer 1

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

$\therefore \frac{k}{h}.m=-1\Rightarrow m=-\frac{h}{k}\because P(h,k)$ lies on tangent

$\therefore k=mh+\sqrt{a^2{m}^2+b^2}$ form equation (ii) & (iii), we get

${(k+\frac{h^2}{k})}^2=\frac{a^2{h}^2}{k^2}+b^2\Rightarrow$ locus is ${(x^2+y^2)}^2=a^2{x}^2+b^2{y}^2$