Question

PQ is the double ordinate of the parabola y²=4ax. Then the locus of its point of trisection is

Answer 1

The equation y^2 = 4ax can be parametrised as:
x = at^2, y = 2at.

If the extremities of the double ordinate are (at^2, 2at) and (at^2, - 2at), then its points of trisection have ordinates y1 and y2 where:
y1 = - 2at + 4at / 3 = 2at / 3
y2 = - 2at + 8at / 3 = - 2at / 3

For both of these points:
y^2 = 4a^2 t^2 / 9
y^2 = (4a / 9)(at^2)
y^2 = 4(a / 9)x.

The locus is the parabola y^2 = 4bx where:
b = a / 9.
hence locus -
y^2 = (4/9) ax