Coordinate Geometry:Locus Based Problem

Question

A rod AB of length l slides with its ends on the coordinate axes.Let O be the origin.The rectangle OAPB is completed.

How to prove the locus of the foot of perpendicular drawn from P onto AB is $x^{2/3}+y^{2/3}=l^{2/3} $ ?

Answer 1

Let $N(x,y)$ be the foot of the perpendicular from $P$ to $AB$, and let angle $OBA=\theta.$

We have $OA=l\sin\theta$ and $PN=l\cos\theta\sin\theta$

Hence $$y=l\sin\theta-PN\cos\theta=l\sin\theta-l\cos^2\theta\sin\theta$$ $$=l\sin^3\theta$$

By similar argument, $x=l\cos^3\theta$.

Therefore, eliminating $\theta$ using $\sin^2\theta+\cos^2\theta=1$, we get the answer