Question

A variable line makes intercept on the coordinate axes.If the length of perpendicular on the line from the origin is the geometric mean of the lengths of intercept, then find the locus of the foot of perpendicular drawn from the origin.

Answer 1

Let the general equation of a line $L1$ be of the form $\frac{x}{a}+\frac{y}{b}=1$. Let $(h,k)$ lie on this line such that a normal line $L2$ passing through this point intersects at the origin. Now we know that $L1$ is perpendicular to $L2$. Thus we get,

$\frac{k}{h}\times \frac{-b}{a}=-1$

Which implies that:

$\frac{k}{h}=\frac{a}{b}$

Also, we know from the information provided in the question - (The length of perpendicular on the line from the origin is the geometric mean of the lengths of intercept) that $ab=a^2+b^2$, $1=\frac{a}{b}+\frac{b}{a}$

From the manipulation we did above we get,

$1=\frac{k}{h}+\frac{h}{k}$

$⟹h^2+k^2=hk$ or $x^2+y^2=xy$

This(ellipse) is the locus of the foot of perpendicular drawn from the origin