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Math Help - Conic Seciton

  1. #1
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    Conic Seciton

    How do I solve x=3y^2+5y-9 ? I am trying to turn it into a conic.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    x= 3(y^2 +\frac{2* 5 y}{6} + \frac{25 }{36}) -\frac{25}{12} - 9

    x= 3(y^2 +\frac{2* 5 y}{6} + {(\frac{5 }{6})}^2) -\frac{25}{12} - 9

    x= 3(y+ \frac{5 }{6})^2 -\frac{133}{12}

    x= 3(y+ \frac{5 }{6})^2 -\frac{133}{12}

    x+\frac{133}{12}= 3(y+ \frac{5 }{6})^2

    (y+ \frac{5 }{6})^2= 4*\frac{1}{3*4} (x+\frac{133}{12})

    Parabola.

    ---------
    Any problem?
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  3. #3
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    Hello, winsome!

    A slightly different path . . .


    x\:=\:3y^2+5y-9

    We have: . x + 9 \:=\:3y^2 + 5y \quad\Rightarrow\quad \frac{x+9}{3} \:=\:y^2 + \frac{5}{3}y


    Complete the square: . \frac{x+9}{3} \: {\color{blue}+ \left(\frac{5}{6}\right)^2} \;=\;y^2 + \frac{5}{3}y \:{\color{blue}+ \left(\frac{5}{6}\right)^2}


    . . . . . . . . . . . . . . . . . . \frac{12x + 133}{36} \;=\;\left(y + \frac{5}{6}\right)^2


    . . . . . . . . . . . . . . . \frac{12}{36}\left(x + \frac{133}{12}\right) \;=\;\left(y+\frac{5}{6}\right)^2


    . . . . . . . . . . . . . . . . . \boxed{\left(y + \frac{5}{6}\right)^2 \;=\;\frac{1}{3}\left(x + \frac{133}{12}\right)}



    This is a "horizontal" parabola, opening to the right: \subset

    . . . . \begin{array}{cc}\text{Vertex:} & \left(-\dfrac{133}{12},\:-\dfrac{5}{6}\right) \\ \\<br />
\text{Focus:} & \left(-11,\:-\dfrac{5}{6}\right) \\ \\<br />
\text{Directrix:} & x \:=\:-\dfrac{67}{6} \end{array}

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