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This is a question I am struggling with as I work through a pure maths book as a hobby:

Prove that the normal to the parabola $y^2=4ax$ at the point $at^2,2at)$ has the equation $y+tx=2at+at^3$. The normals at the points $P(ap^2, 2ap)$ and $Q(aq^2,2aq)$ intersect at the point R. Find the coordinates of R in terms of $(p+q)$ and $pq$. if O is the vertex of the parabola and P and Q are variable points such that $\angle POQ$ is a right angle, find the locus of R; verify that it is a parabola and find the coordinates of the vertex.

For the first part:

$y=2at \rightarrow \frac{dy}{dt}=2a\\ x=at^2 \rightarrow \frac{dx}{dt}=2at\\ \frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}\\$

$\rightarrow$ Gradient of normal = $-t \rightarrow$ equation of normal at $(at^2,2at)$ is:

$\rightarrow y-2at={-t}(x-at^2)\\ \rightarrow y+tx=2at+at^3$

So we then find the point of intersection of the 2 normals at P and Q at point R:

$2ap+ap^3-px=2aq+aq^3-qx$

This eventually gives coordinates of

$\\x=a\{2+(p+q)^2 -pq\}\\ y=-apq(p+q)$

Now I need to use this to find the locus of R. I can find no way get rid of the p's and q's to express y in terms of x. I feel the fact that POQ is a right angle is relevant but cannot see how, unless it is merely to indicate that P and Q are on opposite sides of the x-axis. The book says the answer is $y^2=16a(x-6a)$

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Slope of $OP$ is $$\frac{2ap}{ap^2}=\frac2p$$ Similarly, slope of $OQ$ is $2/q$. If they are perpendicular, it means $$\frac2p\frac2q=-1$$or $pq=-4$. Plug this into your equation for $y$, and write $(p+q)$ in terms of $y$. Then use this expression, and plug it into $x$.

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