Parabola Problem

We take two points on the parabola $A,B$ whose co-ordinates are :

$A=(a{t}_{1}^{2},2a{t}_{1})$

$B=(a{t}_{2}^{2},2a{t}_{2})$

So we know that intersection of their tangents is $P$ where co-ordiantes of $P$ are :

$P=(a{t}_{1}{t}_{2},a({t}_{1}+{t}_{2}))$

To find the co-ordinates of the orthocenter what we do is we write the equation of two altitudes of the triangle $PAB$ and find their points of intersection.

Writing first equation that is altitude from $A$ on $BP$ :

$-{t}_{1}(x-a{t}_{2}^{2})= y-2a{t}_{2}$

$\Rightarrow x{t}_{1}+y=2a{t}_{2}+a{t}_{1}{t}_{2}^{2}$ (i)

Similarly writing second equation that is altitude from $B$ on $AP$ :

$-{t}_{2}(x-a{t}_{1}^{2})= y-2a{t}_{1}$

$\Rightarrow x{t}_{2}+y=2a{t}_{1}+a{t}_{2}{t}_{1}^{2}$ (ii)

Solving them we get :

$x=-a(2+{t}_{1}{t}_{2})$

$y=a({t}_{1}+{t}_{2})(2+{t}_{1}{t}_{2})$

Now it is given the point of intersection of tangents is $(\alpha ,\alpha )$

Comparing it with point P we get :

${t}_{1}{t}_{2}={t}_{1}+{t}_{2}=\alpha$

We assume our orthocenter to be of co-ordinates $(h,k)$

$h=-a(2+\alpha ) , k=a \alpha (2+ \alpha)$

Eliminating $\alpha$ we get :

${h}^{2}+2ah=ak$

Our locus is :

${x}^{2}+2ax=ay$