Given a line with unknown intercepts and finding locus of foot of perpendicular from origin? 
Q) a variable line makes intercepts on coordinate axes . If the length
  of the perpendicular on the line from the origin is the geometric mean
  of the length of intercepts then find the locus of foot of
  perpendicular drawn from origin ?

My Attempt
Given that distance $p^{2}=ab$
And C is a point $(h,k)$ therefore we have 
$h^2 +k^2=ab$ . I also know that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2} \Rightarrow a^2+b^2=ab$ .
Now how do i proceed ...?

Assume that $P=(x_0,y_0)$ is the projection of $O$ on the line. 
The perpendicular to $OP$ through $P$ has equation:
$$ x_0 x + y y_0 = (x_0^2+y_0^2) $$
hence the lengths of the intercepts are given by $\frac{x_0^2+y_0^2}{x_0}$ and $\frac{x_0^2+y_0^2}{y_0}$. 
The problem gives that $OP^2$ equals the product of these lengths, hence $x_0^2+y_0^2=x_0 y_0$ and $P$ belongs to the curve with equation $x^2-xy+y^2=0$, that is an ellipse.
