Find the point P on the parabola y^2 = 4ax such that area bounded by parabola, the X-axis and the tangent at P is equal to that of bounded by the parabola, the X-axis and the normal at P.

Answer 1

Woaahhh! , that looks like a major assignment question.
I will tell you what I would do if I had to actually do this.
let P be (s,t) , but we know that y^2 = 4ax
so t^2 = 4as
s = t^2/(4a)
so we can let P be (t^2/(4a) , t)

from y^2 = 4ax
2y dy/dx = 4a
dy/dx = 2a/y

So the slope of the tangents at P = 2a/t
and the slope of the normal at P = -t/(2a)

equation of tangent at P:
y - t = (2a/t)(x - t^2/(4a) )
we can solve this for x = .... (your job)

we can also solve the parabola for x
x = y^2/(4a)

So we want the area as defined in the first part
This can be done by taking horizontal slices
from y = 0 to y = t
area = ∫(the x of the parabola - x of the tangent) dy from 0 to t
(again, your job)

Now to the other part.
We can also find the equation of the normal, since we have the point P and the slope of the normal
express this as well in terms of x
area of 2nd region
= ∫(x of the normal - x of the parabola) dy from 0 to t

both of these areas will contain the variable t and the constant a
set them equal to each other and solve for t, which will give you the y value of P
Good Luck!