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Thread: Locus&parabola

  1. #1
    Junior Member
    Jul 2009


    P(2ap, ap^2) is a point on the parabola x^2 = 4ay. T is the foot of the perpendicular from P to the x-axis. R is the point where the tangent at P crosses the axis of the parabola. M is the midpoint of RT. Show that the locus of M is x^2 = -2ay

    I'm having SO much trouble on this question, ive tried so many ways but they never come to that equation
    I worked out that the co-ordinates of point R are (0,y) and T (2ap,0)
    I then went and figured out the m(RT) and then the equation of RT.
    m(RT) = -y/2ap
    Then I worked out the midpoint of RT, M which was (ap, y/2)
    but i am stuck on what to do next, i tried a few different approaches and they all didnt work, am i even going in the right direction?? T.T""
    please helpp me !
    thankyou =)
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    We have $\displaystyle P(2ap,ap^2), \ T(2ap,0)$

    The equation of the tangent is $\displaystyle x_0x=2a(y+y_0)$, where $\displaystyle (x_0,y_0)$ are the coordinates of P.

    Therefore, $\displaystyle 2apx=2a(y+ap^2)$

    $\displaystyle x=0\Rightarrow y=-ap^2\Rightarrow R(0,-ap^2)$

    Then $\displaystyle M\left(ap,-\frac{ap^2}{2}\right)$

    We have

    $\displaystyle \left\{\begin{array}{ll}x=ap\\y=-\frac{ap^2}{2}\end{array}\right.$

    From the first equation $\displaystyle p=\frac{x}{a}$. Replace p in the second: $\displaystyle y=-\frac{x^2}{2a}\Rightarrow x^2=-2ay$
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