Let the given points be A(x1, y1) = \(\big(at\frac{2}{1}, 2at_1\big)\),
B(x2, y2) = \(\big(at\frac{2}{2}, 2at_2\big)\), and C(x3, y3) = (a, 0)
∴ Area of ∆ABC
= \(\frac{1}{2}\)|x1 (y2 − y3) + x2 (y3 − y1) + x3 (y1 − y2)|
If t1t2 = – 1, then t1t2 + 1 = 0
∴ Area of ∆ABC = 0
Hence, the given points are collinear.