Question

The perpendicular OY is drawn from the centre O of the ellipse x²/a² + y²/b² = 1 to any tangent. Prove that the locus of Y is (x² + y² )² = a²x² + b² y².

Answer 1

Let p(x,y) be the foot of perpendicular to a tangent

$y=mx\pm \sqrt{a^2m^2+b^2}\cdots (1)\\$ from Centre

slope of OY = $\frac{-1}{m}\\$

Equation of OY is

$y=\frac{-1x}{m} \implies m=\frac{-x}{y} \cdots (2)$

p(x,y) lies on tangent,

$y=\frac{-x}{y}.x\pm\sqrt{a^2(\frac{-x}{y})^2+b^2}\\ y=\frac{-x}{y}.x\pm\sqrt{a^2\frac{x^2}{y^2}+b^2}\\ y=\frac{-x}{y}\pm\sqrt{\frac{a^2x^2+b^2y^2}{y^2}}\\ y^2+x^2=\pm\sqrt{a^2x^2+b^2y^2}\\ (x^2+y^2)^2=x^2a^2+b^2y^2.$