Find the locus of the foot of perpendiculars drawn from the vertex on a variable tangent to the parabola $y^2=4ax$

Question

The tangent will be $$y=mx+\frac am$$

The line perpendicular to this will be $$y=-\frac 1m x$$

Solving these equations $$x=-\frac{a}{1+m^2}$$ And $$y=\frac{a}{m(1+m^2)}$$

I am not able to eliminate m to obtain the locus. How should I proceed?

Answer 1

$$\dfrac xy=\cdots=-m$$

Replace the value of $m$ in anyone of the last two relations to eliminate the foreign variable $m$

Answer 2

Yes, take and eliminate $m$ from them $$x=-\frac{a}{1+m^2}~~~(1), y=\frac{a}{m(1+m^2)}~~~(2)$$ dividing them you get $m=-x/y$ put it in (1) to get the required locus as $$x(x^2+y^2)=-ay^2.$$