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Math Help - Line Tangent To Parabola's

  1. #1
    Member Rimas's Avatar
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    Line Tangent To Parabola's

    Just Needed Help on this:
    A line is Tangent to a parabola Y^2=4ax at a point P(not the Vertex) and intersects the axis of this parabola at a point Q. Prove that the line segment PQ is bisected by the line which is tangent to the parabola at its vertex.
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  2. #2
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    Hello, Rimas!

    A line is tangent to a parabola y^2=4ax at a point P (not the vertex)
    and intersects the axis of this parabola at a point Q.
    Prove that the line segment PQ is bisected by the line
    which is tangent to the parabola at its vertex.

    If they made a sketch, they could have stated the problem more clearly.
    Code:
                    | 
                    |      P   o     *
                    |      o *
                    |  o* (p,q)
                   o| *
               o    |*
           o        |
      -o- - - - - - * - - - - - - - - -
       Q            |
                    |*
                    | *
                    |   *
                    |
    The parabola is "horizontal", vertex at (0,0), opens to the right.
    The tangent at P(p,q) has its x-intercept at Q.
    Prove that PQ is bisected by the y-axis.


    We have: . y^2 = 4ax \quad\Rightarrow\quad 2yy' = 4a \quad\Rightarrow\quad y' = \frac{2a}{y}

    At P(p,q), the slope of the tangent is: .  m = \frac{2a}{q}

    The equation of the tangent is: . y - q \:=\:\frac{2a}{q}(x-p) \quad\Rightarrow\quad y \:=\:\frac{2a}{q}x + \frac{q^2-2ap}{q}

    Since q^2 = 4ap, we have: . y \:=\:\frac{2a}{y}x + \frac{4ap - 2ap}{q} \quad\Rightarrow\quad y \:=\:\frac{2a}{q}x + \frac{2ap}{q}


    Let y = 0, the x-intercept is: . Q(-p,0)


    Let x = 0, the y-intercept is: . M\left(0, \frac{2ap}{q}\right) .[1]

    Since q^2 = 4ap, then 2ap = \frac{q^2}{2}

    . . Substitute into [1]: . M\left(0, \frac{\frac{q^2}{2}}{q}\right) \:=\:M\left(0,\frac{q}{2}\right)


    And M\left(0,\frac{q}{2}\right) is indeed the midpoint of P(p,q) and Q(-p,0)

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