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Thread: 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 $\displaystyle Y^2=4ax$ 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 $\displaystyle y^2=4ax$ at a point $\displaystyle P$ (not the vertex)
    and intersects the axis of this parabola at a point $\displaystyle Q.$
    Prove that the line segment $\displaystyle PQ$ 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 $\displaystyle P(p,q)$ has its x-intercept at $\displaystyle Q.$
    Prove that $\displaystyle PQ$ is bisected by the y-axis.


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

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

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

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


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


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

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

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


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

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