Parabola Ques

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.

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  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!

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