When the normals at two points on a parabola meet at a point on that parabola Question

Consider the parabola with equation $y^2 = 4 x$, parameterized by $(x,y) = (t^2,2t)$. If the normals at points corresponding to distinct parameter values $t_1$ and $t_2$ meet at a point on the parabola, find the relation between $t_1$ and $t_2$.
(Note from Blue: It's possible, even likely, that the question intends the intersection to be a distinct third point on the parabola.)

My Attempt:
Let the points be $(t_1^2,2t_1)$ and  $(t_2^2,2t_2)$.
By taking the derivative at $t_1$ and $t_2$, I get the slopes of the normals as $-t_1$ and $-t_2$.
I then used point-slope form to get the equation of the two normals, and found the $x-$coordinate of intersection as $$x=t_1^2+t_2^2+t_1t_2+2$$
I am stuck over here.
 A: So the points are $(t_1^2,2t_1)$ and $(t_2^2,2t_12)$
The slopes of the normals are $-t_1$ and $-t_2$ respectively.
The equations of the normal lines are
\begin{align}
   t_1x + y &= t_1^3+2t_1 \\
   t_2x + y &= t_2^3+2t_2 \\
\hline
   (t_2-t_1)x &= (t_2^3-t_1^3) + 2(t_2-t_1) \\
   x &= t_2^2 + t_2t_1 + t_1^2 + 2 \\
   y &= t_1^3+2t_1 -t_1(t_2^2 + t_2t_1 + t_1^2 + 2) \\
   y &= -t_1^2t_2 - t_1t_2^2
\end{align}
Since the point (x,y) needs to also be on the parabola, we need
\begin{align}
   y^2 &= 4x \\
   (-t_1^2t_2 - t_1t_2^2)^2 &= 4(t_2^2 + t_2t_1 + t_1^2 + 2) \\
   (t_1 t_2 - 2) (t_1^2 + t_2 t_1 + 2) (t_2^2 + t_1 t_2 + 2) &= 0
\end{align}
So $t_1t_2=2$ or $t_1(t_1+t_2)=-2$ or $t_2(t_1+t_2)=-2$
Note that these are pairwise incompatible (except for the last two, but that would mean both points are necessarily the same - the points cannot be vertically opposite each other); one parameter would become imaginary.
Added because of comment by 'Blue 6'.
We have $(x,y)=((t_1 + t_2)^2 + 2 - t_1t_2, -t_1t_2(t_1+t_2))$
If $t_1t_2=2$, $(x,y)=((t_1 + t_2)^2, -2(t_1+t_2))$.
If $t_1(t_1+t_2)=-2$, then $(x,y)=(t_2^2, 2t_2)$, that is, the normal line through the point $(t_1^2, 2t_1)$ on the parabola contains the point $(t_2^2, 2t_2)$ on the parabola.
If $t_2(t_1+t_2)=-2$, then $(x,y)=(t_1^2, 2t_1)$, that is, the normal line through the point $(t_2^2, 2t_2)$ on the parabola contains the point $(t_1^2, 2t_1)$ on the parabola.
So,if the normal lines to the parabola $y^2=4x$ at the distinct points $(t_1^2,2t_1)$ and $(t_2^2,2t_12)$ are to pass through a distinct third point on the parabola, then we need  $t_1t_2=2$.
A: Hint: Also solve for the $y$-coordinates of the point of intersection. The equation of the parabola is $y^2=4x$. Substitute the point of intersection into the equation of the parabola.
