Question:
The point P $(2p,2/p)$ lies on the rectangular hyperbola C with equation $xy = 4$.
(a) Find the equation of the normal to C at P.
The normal at P meets C again at the point Q. The mid-point of PQ is the point M.
(b) Find, in cartesian form, an equation of the locus of M as p varies.
My attempt:
I have attempted this so far and have found the following equations for the normals, midpoints etc. 
The second equation is the equation of the normal to the hyperbola $xy=4$.
I have then found the midpoint of the PQ to be M $([p^4-1]/p^3 , [1-p^4]/p)$ and have then attempted to eliminate $p$ to find the cartesian equation for the locus of points but I am left with a $p^2$ which I do not know how to get rid of.
The equation of the locus that I come up with is $y=-p^2x$ which comes by equating the $p^4-1$ from both $x$ and $y$ which leaves me with $py = -p^3x$ and so cancelling a $p$ from each side gives me the equation $y= -p^2x$ which is where I become stuck. My guess from the graph of the locus of the midpoint is that the function is some sort of cubic function but that is only a guess.
You have already the equation of the normal at $P(2p,\frac 2p)$ which is $$y-\frac 2p=p^2(x-2p)$$
Let this meet the curve again at $Q(2q,\frac 2q)$
Then, $$\frac 2q-\frac 2p=p^2(2q-2p)$$ We can factor out $(p-q)\neq0$ $$\Rightarrow q=-\frac{1}{p^3}$$ The midpoint of $PQ$ has coordinates $(x,y)$ given by $$x=p+q, y=\frac 1p+\frac 1q=\frac {p+q}{pq}$$ Hence $$pq=\frac xy=-\frac{1}{p^2}\Rightarrow p^2=-\frac yx\Rightarrow q^2=\frac{1}{p^6}=-\frac{x^3}{y^3}$$
Finally we can use the identity $$(p+q)^2=p^2+q^2+2pq$$ and we get $$x^2=-\frac yx-\frac{x^3}{y^3}+\frac {2x}{y}$$ This simplifies as $$x^3y^3+(x^2-y^2)^2=0$$