Given question:
$P, Q, R$ are three points on a parabola and the chord $PQ$ cuts the diameter through $R$ in $V$. Ordinates $PM$ and $QN$ are drawn to this diameter. Prove that $RM.RN = RV^2$
What I did: I represented the three as parametric points with parameters $t_1, t_2, t_3$ on parabola $y^2 = 4ax$. I found the equation of the chord and then its intersection V with the diameter through R. I then dropped perpendiculars from P and Q to the diameters and took their feet as M and N. But then this is the outcome $$RM = a(t_1^2-t_3^2)$$ $$RN = a(t_2^2-t_3^2)$$ $$RV = -a(t_1-t_3)(t_2-t_3)$$
Which doesn't seem matching with what's been asked to prove. Where am I going wrong or is the question itself wrong?
In a parabola the abscissa is proportional to the square or the related ordinate, that is: $$ RM=kPM^2,\quad RN=kQN^2. $$ On the other hand, by similar triangles we have: $$ \begin{align} VM/VN &= PM/QN \\ (VM+VN)/VN &= (PM+QN)/QN \\ (RN-RM)/VN &= (PM+QN)/QN \\ k(QN^2-PM^2)/VN &= (PM+QN)/QN \\ k(QN-PM)QN &= VN \\ kQN^2-kPM\cdot QN &= VN \\ RN-kPM\cdot QN &= VN \\ RN-VN &= kPM\cdot QN \\ RV &= kPM\cdot QN. \end{align} $$ Hence: $$ RV^2 = kPM^2\cdot kQN^2 =RM\cdot RN. $$