Chapter 9.3, Problem 21WE

To determine

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

Answer to Problem 21WE

The required vertex, focus, directrix and axis of symmetry of the parabola are $\left(-2,1\right),\left(-2,-1\right),y=3$ and $\left(x=-2\right)$ respectively.

Explanation of Solution

Given information: The equation of the parabola is $x^2+8y+4x-4=0$ .

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

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

  $\begin{array}{l}{x}^2+8y+4x-4=0\\ x^2+4x+4+8y-4=4\\ x^2+4x+4+8y-8=0\\ {\left(x+2\right)}^2+8y-8=0\\ 8y-8=-{\left(x-2\right)}^2\\ 8\left(y-8\right)=-{\left(x-2\right)}^2\\ y-1=-\left(\frac{1}{8}\right){\left(x-\left(-2\right)\right)}^2\end{array}$

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

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

Now,

The vertex are $\left(h,k\right)=\left(-2,1\right)$

The focus is $\left(h,k+c\right)=\left(-2,\left(1-2\right)\right)=\left(-2,-1\right)$

The directrix is $y=k-c$

  $\begin{array}{l}=1-\left(-2\right)\\ \therefore y=3\end{array}$

The symmetric axis is $x=h=-2$

Now, graph the given parabola,