The orthocenter of a triangle formed by tangents to the parabola $y^2=4ax$ at  $t_1,t_2,t_3$ lies on 

$$\begin{gathered} \text{ Tangents at }P\left(t_1\right),Q\left(t_2\right)\text{ and }R\left(t_3\right)\text{ are } \\ \begin{array}{r}{\mathrm{t}}_1\mathrm{y}=\mathrm{x}+{\mathrm{att}}_1^2\cdots -\left(1\right)\\ {\mathrm{t}}_2\mathrm{y}=\mathrm{x}+{\mathrm{att}}_2^2\cdots -\left(2\right)\\ {\mathrm{t}}_3\mathrm{y}=\mathrm{x}+{\mathrm{att}}_3^2\cdots -\left(3\right)\end{array} \\ \text{ Points of intersection of these tangents are } \\ \begin{array}{r}A\left(a{t}_1{t}_2,a\left(t_1+t_2\right)\right)\\ B\left(a{t}_2{t}_3,a\left(t_2+t_3\right)\right)\\ C\left(a{t}_3{t}_1,a\left(t_3+t_1\right)\right)\end{array} \end{gathered}$$
$$\begin{gathered} \text{Altitudes from vertex }A and B are \\ y-a\left(t_1+t_2\right)=-t_3\left(x-a{t}_1{t}_2\right) \\ y-a\left(t_2+t_3\right)=-t_1\left(x-a{t}_2{t}_3\right) \\ solve \\ Orthocentre=\left(-a,a\left(t_1+t_2+t_3\right)+a{t}_1{t}_2{t}_3\right) \\ \text{ Which is on the directrix of the parabola. } \end{gathered}$$