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I need help with parametric equations, locus problems, tell me how to solve all questions if you can. Need help with one below:

Tangents are drawn to a parabola x^2=4y
from an external point A(x1,y1) touching the parabola at P and Q
(a) Prove that the mid point, M, is the point (x1,0.5 x1^2-y1)
(b) if A moves along the straight line y=x-1 find the equation of the locus M

  • math -

    For ease of typing I am going to let
    A(x1, y1) = A(a,b)

    Let P(x,y) be the point of contact
    slope of AP = (b-y)/(a-x)

    but slope of tangent by Calculus ,(differentiate x^2 = 4y )
    = x/2
    so (b-y)/(a-x) = x/2
    2b - 2y = ax - x^2
    2b - 2(x^2/4) = ax - x^2
    x^2 - 2ax + 4b = 0
    by quadratic equation
    x = (2a ± √(4a^2 - 16b)/2
    = (a ± √(a^2 - 4b)

    so P is ( (a + √(a^2 - 4b) , [(a + √(a^2 - 4b)]^2/4 )
    and Q is ( (a - √(a^2 - 4b) , [(a - √(a^2 - 4b)]/4)

    now remember how to find the midpoint?
    for the x coordinate, add the 2 x's of the endpoints and divide by 2
    clearly that would give us a
    for the y coordinate, I will not type it all out, but when you expand the above y values of the endpoints and then add them up , the middle terms drop out and I got
    y = (2a^2 - 4b)/4
    = (a^2/2 - b)

    M is (a, a^2/2 - b) or in your original version
    M is (x1, x1^2/2 - y1) which is what we were supposed to prove.

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