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Math Help - HELPfocus of parabola!!

  1. #1
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    HELPfocus of parabola!!

    could someone please walk through the steps of solving this problem....?

    find the focus of the conic section y=-3x^2-12x-5
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  2. #2
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    Hello, emetty90!

    Find the focus of the parabola: . y\:=\:-3x^2-12x-5

    We must get the equation into the form: . (x-h)^2 \:=\:4p(y-k)

    Then the vertex is at (h,k) and p is the directed distance from the vertex to the focus.


    We have: . 3x^2 + 12x \:=\:\text{-}y - 5

    Divide by 3: . x^2 + 4x \:=\:\text{-}\tfrac{1}{3}y - \tfrac{5}{3}

    Complete the square: . x^2 + 4x {\color{red}\;+\; 4} \:=\:\text{-}\tfrac{1}{3}y - \tfrac{5}{3} {\color{red}\;+\; 4} \quad\Rightarrow\quad (x+2)^2 \:=\:\text{-}\tfrac{1}{3}y+\tfrac{7}{3}

    . . And we have: . (x+2)^2 \;=\;\text{-}\tfrac{1}{3}(y - 7)


    The vertex is: . V(-2,7) . . . and the parabola opens downward.

    Since 4p = \text{-}\tfrac{1}{3} \quad\Rightarrow\quad p \:=\:\text{-}\tfrac{1}{12} . . . The focus is \tfrac{1}{12} unit below the vertex.


    Therefore, the focus is: . \left(\text{-}2,\,7\,\text{-}\,\tfrac{1}{12}\right) \;=\;\left(\text{-}2,\frac{83}{12}\right)

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