The parabola y2 = 4ax, where a > 0, and the rectangular hyperbola xy = C2, where
C > 0, intersect at right angles. Show that the tangent and normal to either curve at the
point of intersection meet the x-axis at T and N where TN = 2pa, where p is an integer
to be determined.

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  1. from y^2 = 4ax
    2y dy/dx = 4a
    dy/dx = 2a/y ---- >slope of tangent to parabola

    from xy = c^2
    xdy/dx + y = 0
    dy/dx = -y/x -----> slope of tangent of hyperbola

    but we are told that they intersect at right angles, so the tangents must be perpendicular, making
    2a/y = x/y
    x = 2a
    sub into xy=c^2
    y = c^2/(2a)

    so the intersect at (2a, c^2/(2a)
    (Now it gets messy)
    at parabola, slope = 2a/(c^2/2a) = 4a^2/c^2

    equation of tangent:
    y - c^2/(2a) = (4a^2/c^2)(x-2a)
    yc^2 - c^4/(2a) = 4a^2x - 8a^3
    at the x-axis, y = 0
    - c^4/(2a) = 4a^2x - 8a^3
    4a^2x = 8a^3 - c^4/(2a)
    x = 2a - c^4/(8a^3) -----> the x-intercept, call it T

    at the hyperbola, slope =
    slope = -y/x = (-c^2/(2a))/(2a) = -c^4/(4a^2)
    notice that this is the negative reciprocal of the slope at the parabola, so far so good!

    equation of tangent:
    (y - c^2/(2a)) = (-c^2/(4a^2)) (x - 2a)
    4a^2y - 2ac^2 = -c^2x + 2ac^2
    again at the x-intercept , let y = 0
    - 2ac^2 = -c^2x + 2ac^2
    c^2 x = 4ac^2
    x = 4a ------------> the other xintercept, call it N

    so TN = 4a - (2a - c^4/(8a^3)) = 2a + c^4/(8a^3)
    but TN = 2pa
    2a + c^4/(8a^3) = 2pa
    p =1 + c^4/(16a^4)

    ????????? What do you think????

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