# How do you find the vertex, focus, and directrix of the following parabola and graph the equation y^2-8y+8x+16=0?

May 4, 2017

#### Answer:

Vertex is $\left(0 , 4\right)$, focus is $\left(- 2 , 4\right)$ and directrix is $x = 2$

#### Explanation:

${y}^{2} - 8 y + 8 x + 16 = 0$

$\Leftrightarrow 8 x = - {y}^{2} + 8 y - 16$

or $8 x = - \left({y}^{2} - 8 y + 16\right)$

or $8 x = - {\left(y - 4\right)}^{2}$

or $x = - \frac{1}{8} {\left(y - 4\right)}^{2}$

Hence vertex is $\left(0 , 4\right)$ and axis of symmetry is $y = 4$

As coefficient of ${\left(y - 4\right)}^{2}$ is negative, parabola opens on left-hand side and hence focus will be to the left of vertex and directrix on the right.

As this coefficient is $\frac{1}{8}$, equating it to $\frac{1}{4 p}$, we get $p = 2$ and focus is $\left(- 2 , 4\right)$ and directrix is $x = 2$

graph{(y^2-8y+8x+16)(x-2)(y-4)((x+2)^2+(y-4)^2-0.04)=0 [-20.25, 19.75, -6.36, 13.64]}