Prove that the equation of the chord joining the points $P(cp,\displaystyle\frac{c}{p})$ and $Q(cq,\displaystyle\frac{c}{q})$ on the rectangular hyperbola $xy=c^2$ is $pqy+x=c(p+q)$.
So, I've found that the gradient of the chord $PQ$ is $\displaystyle\frac{-1}{pq}$ and managed to prove that the equation of the chord is $pqy+x=c(p+q)$.
It is given that $PQ$ subtends a right angle to the tangent at the point $R(cr,\displaystyle\frac{c}{r})$ on the curve, prove that the midpoint of PQ lies on the straight line $y+r^2x=0$.
I've obtained that the midpoint of $PQ$ is $\displaystyle\big(\frac{cp+cq}{2},\frac{\frac{c}{p}+\frac{c}{q}}{2}\Big)$
Substitute $y=\displaystyle\frac{\frac{c}{p}+\frac{c}{q}}{2}$ into $y+r^2x=0$,
$\displaystyle\frac{\frac{c}{p}+\frac{c}{q}}{2}+r^2x=0$
$cp+cq=-2pqr^2x$
By substituting value of y into the equation, I expected to obtain the value of x, but how do I get rid of $r$?
How do I proceed? And how should I use the information given about the tangent at point R?
The gradient of $PR$ $$=\dfrac{c/p-c/r}{cp-cr}=-\dfrac1{pr}$$
Similarly, the gradient of $QR$ $$=-\dfrac1{qr}$$
So, we have $\dfrac1{pqr^2}=-1\iff pq=?$
If $M(h,k)$ is the midpoint $PQ,$
$$2h=c(p+q)$$
$$2k=\dfrac cp+\dfrac cq=?$$
$$\dfrac{2h}{2k}=\cdots=pq=-\dfrac1{r^2}$$