This is a question I found: P $(t_1,t_1^2) $ and Q $(t_2,t_2^2)$ are two points on a parabola such that PQ subtends a right angle at the vertex. What is the locus of points of intersection of normals at P and Q? The answer given is $x^2 = 2(2y-3)$.
Now I found that $ t_1t_2 = -1$. The tangents at P and Q will be given by the equations $$ y = 2xt_1 - t_1^2 $$ and $$ y = 2xt_2 - t_2^2 $$. I concluded that the normals will have the negative reciprocal as slope and are given by $$ y = -x/2t_1 +c1 $$ and $$ y = -x/2t_2 +c2 $$. On substituting $(t_1,t_1^2) $ in first normal equation I get $c_1 = (2x^2t_1+1/2t_1)$. This doesn't seem to be right. Is my method right? Also, how do I proceed to find the locus of points of intersection of these normals?