The normal at $A$ and $B$ on the parabola $y^2=4ax$ meet the parabola at $C$ on same parabola.
Then locus of orthocenter of $\triangle ABC$
Attempt Let $A(at^2_{1},2at_{1})$ and $B(at^2_{2},2at_{2})$. Then equation of Normal $A$ and $B$ is
$y=-t_{1}x+2at_{1}+at^3_{1}$ and $y=-t_{2}x+2at_{2}+at_{2}^3$
So coordinate of point $C\bigg(a(2+t^2_{1}+t^2_{2}+t_{1}t_{2}),at_{2}(t_{1}+t_{2})\bigg)$
could some help me how to solve it, thanks
You must start with point $C=(y_C^2/4a,y_C)$: the normal at a generic point $(y^2/4a,y)$ on the parabola passes through $C$ if: $$ y-y_C=-{y\over2a}{y^2-y_C^2\over4a}, \quad\hbox{that is:}\quad y={-y_C\pm\sqrt{y_C^2-32a^2}\over2}. $$ These two solutions above give then the coordinates of $A$ and $B$ (provided of course $y_C^2\ge32a^2$): $$ y_A={-y_C+\sqrt{y_C^2-32a^2}\over2},\quad y_B={-y_C-\sqrt{y_C^2-32a^2}\over2},\quad $$ and of course $x_A=y_A^2/4a$, $x_B=y_B^2/4a$.
Now that you know the coordinates of $A$, $B$ and $C$ (as a function of $y_C$) I think you can go on by yourself to find the orthocenter of triangle $ABC$, whose locus should lie on another parabola.
EDIT.
If $O=(x_O,y_O)$ is the orthocenter, a straightforward computation gives in fact:
$$ x_O=x_C-6a,\quad y_O=-{y_C\over2}, $$ and $O$ lies then on the parabola of equation $y^2=ax+6a^2$.