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Thread: Find equation of the parabola

  1. #1
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    Find equation of the parabola

    If the axis is parallel to x axis, Latus Rectum = 1 and passing through (3,1) and (-5,5)

    Help please. No idea. The axis parallel to x axis made it confusing.
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  2. #2
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    Hello reiward
    Quote Originally Posted by reiward View Post
    If the axis is parallel to x axis, Latus Rectum = 1 and passing through (3,1) and (-5,5)

    Help please. No idea. The axis parallel to x axis made it confusing.
    The 'standard' parabola with axis parallel to the $\displaystyle x$-axis has equation
    $\displaystyle y^2 = 4ax$
    and if we move the vertex from the origin to the point $\displaystyle (h,k)$, this equation becomes:
    $\displaystyle (y-k)^2 = 4a(x-h)$
    Now the latus rectum has length $\displaystyle 4a$. So here $\displaystyle 4a = 1$, and the equation now becomes:
    $\displaystyle (y-k)^2 = (x-h)$
    Plug in the values of $\displaystyle x$ and $\displaystyle y$ at the two points you're given:
    $\displaystyle \left\{\begin{array}{l}
    (1-k)^2=3-h\\
    (5-k)^2=-5-h\\
    \end{array}\right.$

    $\displaystyle \Rightarrow\left\{\begin{array}{l}
    k=4\\
    h=-6\\
    \end{array}\right.$
    So the equation is:
    $\displaystyle (y-4)^2 = x+6$
    which can be re-written as:
    $\displaystyle x=y^2-8y+10$
    if you like.

    Grandad
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