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?
Equation of first normal:
$$y-t^2=-\frac{1}{2t}(x-t) \tag{1}$$
Equation of second normal:
$$y-\frac{1}{t^2}=\frac{t}{2}\left( x+\frac{1}{t} \right) \tag{2}$$
$(2)-(1)$,
\begin{align} \left( t^2-\frac{1}{t^2} \right) &= \left( \frac{t}{2}+\frac{1}{2t} \right) x \\ x &= 2 \left( t-\frac{1}{t} \right) \\ y &= t^2+\frac{1}{t^2}-\frac{1}{2} \\ &= \left( t-\frac{1}{t} \right)^2+\frac{3}{2} \\ &= \frac{x^2}{4}+\frac{3}{2} \\ x^2 &= 2(2y-3) \end{align}