I'm quite lost solving the below problem.
P, Q, R and S are four points on the hyperbola $x = ct$, $y = \frac{c}{t}$ with parameters p, q, r, and s respectiveley. Prove that, if the chord PQ is perpendicular to the chord RS, then $pqrs = -1$.
I know that $\frac{dy}{dx} = {\frac{dy}{dt}} / \frac{dx}{dt}$, but don't see how that fits into the picture, except that when multiplying the gradient of two perpendicular lines, the result is $-1$.
Hint: Think about the vectors $PQ=P-Q$ and $RS=R-S$.
Answer: We have that $P=(cp,\frac cp)$, $Q=(cq,\frac cq)$, $R=(cr,\frac cr)$ and $S=(cs,\frac cs)$. Then $$PQ=(cp-cq, \frac cp - \frac cq)= (c(p-q), c\frac{q-p}{pq})$$ and $$RS=(cr-cs, \frac cr - \frac cs)= (c(r-s), c\frac{s-r}{rs}).$$ Since $PQ\perp RS$, we have that $PQ\cdot RS=0$, i.e. $$c^2(p-q)(r-s)+c^2\frac{q-p}{pq}\frac{s-r}{rs}=c^2(p-q)(r-s)(1+\frac{1}{pqrs})=0.$$ Since $c\neq 0$, $p\neq q $ and $r\neq s$, it follows that $$1+\frac{1}{pqrs}=0,$$ that is, $$pqrs=-1.$$