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 ...?

Let the general equation of a line (L1) be of the form x/a+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,
(k/h)*(-b/a)=-1
Which implies that:
k/h=a/b 
Also, we know from the information provided in the question that 
ab=a^2+b^2, 
1=a/b+b/a
From the manipulation we did above we get,
1=k/h+h/k
which implies that 
h^2+k^2=hk or x^2+y^2=xy.
