I am struggling with the last part of this question.

Prove that the tangent at the point $(at^2,2at)$ on the parabola $y^2=4ax$ has the equation $ty=x+at^2$. Find, in their simplest form, the coordinates of T, the point of intersection of the tangents at the points $P(ap^2,2ap)$ and $Q(aq^2,2aq)$ on the parabola $y^2=4ax$. If PQ is of constant length $d$, show that T lies on the curve whose equation is:

$(y^2-4ax)(y^2+4a^2)=a^2d^2$

My workings for the first parts are as follows:

$x=at^2; y=2at$

$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{1}{t}$

$\rightarrow y-2at=\frac{1}{t}(x-at^2)\rightarrow ty=x+at^2$ (1)

Equation of tangent at P: $py=x+ap^2$

Equation of tangent at Q: $qy=x+aq^2$

At T: $\frac{x+ap^2}{p} = \frac{x+aq^2}{q} \rightarrow qx = ap^2q= px+aq^2p \rightarrow x=apq$

from (1) $y=\frac{x+ap^2}{p} \rightarrow y=a(p+q)$

So coordinates of T are $apq, a(p+q$

Then comes this part which I cannot solve:If PQ is of constant length $d$, show that T lies on the curve whose equation is:

$(y^2-4ax)(y^2+4a^2)=a^2d^2$

I am assuming the calculations I have already done will help me but I cannot see the link.