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 intercepts then find the locus of the foot of perpendicular draws from the origin.

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.

Regards

Arun (askIITians forum expert)