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Thread: parabola

  1. #1
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    parabola

    In need to write an equation for the parabola whose vertex is (4, 3), axis of symmetry is y = 3 and the measure of the latus rectum is 4; a > 0. I also need to draw this parabola.
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  2. #2
    Senior Member vincisonfire's Avatar
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    A parabola is the set of all points in the plane equidistant from a given line.
    $\displaystyle \sqrt{(y-12)^2+(x-3)^2}=y+4 $
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  3. #3
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    Hello, 14041471!

    Write an equation for the parabola whose vertex is (4, 3),
    axis of symmetry is $\displaystyle y = 3$ and the measure of the latus rectum is 4; $\displaystyle a > 0.$
    I also need to draw this parabola.

    You might start with a sketch . . .
    Code:
            |
            |                 *
            |             *
            |          *  :
            |  (4,3) *  p :
          - + - - - o - - o - -
            |      V *    :F
         ---+----------*--:------
            |             * 
            |                 *
            |
    We're expected to know the following . . .

    Since the axis of symmtetry is horizontal, the parabola opens "sideways."
    Since $\displaystyle a > 0$, the parabola opens to the right.

    The general form is: .$\displaystyle (y-k)^2 \:=\:4p(x-h)$
    . . where $\displaystyle (h,k)$ is the vertex,
    . . and $\displaystyle p$ is the focal distance (distance from vertex to focus).

    Fact: the length of the latus rectum is $\displaystyle 4p.$
    . . . . We have: .$\displaystyle 4p \:=\:4\quad\Rightarrow\quad p \:=\:1$


    Since the vertex is: .$\displaystyle (h,k) = (4,3)$

    . . the equation is: .$\displaystyle (y-3)^2 \;=\;4(x-4)$

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  4. #4
    Senior Member vincisonfire's Avatar
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    ! Wrong direction for me. May I swap x's and y's?
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