The ordinates of the feet of three normals to the parabola `y^2=4ax` from the point (6a, 0) are

Explanation

We need to draw three normals from the point `(6a,0)` on the parabola `y^2 = 4ax`
`Eqn.` of normal in parametric form:
`y = -tx + 2at + at^3`
This will pass from `(6a,0)`, so it will satisfy the above equation.
`rArr 0 = -t*6a + 2at + at^3`
`rArr at^3 - 4at = 0`
`rArr at(t^2 - 4) = 0`
`rArr t = 0 or t = pm 2`
...