1. ## Conic Seciton

How do I solve x=3y^2+5y-9 ? I am trying to turn it into a conic.

2. $x= 3(y^2 +\frac{2* 5 y}{6} + \frac{25 }{36}) -\frac{25}{12} - 9$

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

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

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

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

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

Parabola.

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Any problem?

3. 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: $\subset$

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