Chapter 9.3, Problem 22WE

To determine

To calculate: The vertex, focus, directrix, and axis of symmetry.

Answer to Problem 22WE

The required vertex, focus, directrix and axis of symmetry of the parabola are $\left(2,-3\right),\left(0,-3\right),x=6$ and $\left(y=-3\right)$ respectively.

Explanation of Solution

Given information: The equation of the parabola is $y^2+6y+8x-7=0$ .

Formula used: $x-h=\frac{1}{4c}{\left(y-k\right)}^2$

Calculation: Rewrite the given equation in to general form of equation,

  $\begin{array}{l}{y}^2+6y+8x-7=0\\ y^2+6y=-8x+7\\ y^2+6y+9=-8x+7+9\\ {\left(y+3\right)}^2=-8\left(x-2\right)\end{array}$

Now, compare ${\left(y+3\right)}^2=-8\left(x-2\right)$ to equation $x-h=\frac{1}{4c}{\left(y-k\right)}^2$ , we get,

  $\begin{array}{c}h=2\\ k=-3\\ \begin{array}{l}y=k\\ \therefore y=-3\end{array}\\ \begin{array}{l}\frac{1}{4c}=-\frac{1}{8}\\ c=-2\end{array}\end{array}$

Now,

The vertex is $\left(h,k\right)=\left(2,-3\right)$

The focus is $\left(h+c,k\right)=\left(0,-3\right)$

The directrix is $x=h-c$

  $\begin{array}{l}=2-\left(-4\right)\\ \therefore x=6\end{array}$

The axis of symmetry is $y=k=-3$

Now, graph the given parabola,