# Find the locus of the point of intersection of 2 normals to a parabola

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 $$y^2=16a(x-6a)$$

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 $$x$$.