If the normal to the parabola y2 = 4ax at P(at1t2, 2at1) meets it again at the point t2, then we have t2 = – t1 – 2/t1
If the abscissa and the ordinates of P be equal, then at12 = 2at1
⇒ t1 = 2 (rejecting t1 = 0) ? t2 = – 2 – 1 = – 3
The co-ordinates of P and Q are therefore (4a, 4a) and (9a, – 6a) respectively.
The focus is the point S (a, 0).
Slope of PS = 3/4 and slope of QS = – 3/4.
⇒ ∠PSQ = right angle. Hence the result.