Locus of the point of intersection of the normals to the parabola y2=16x which are at right angles is

Let two perpendicular normals to the parabola y2=16xbe y=mx−8m−4m3  and  y=−1mx+8m+4m3Solving we get x=4m−1m2+12y=−4m−1mEliminating m we get the required locus asy2=4(x−12).