Chapter 9.3, Problem 7ST

To determine

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

Answer to Problem 7ST

The required vertex, focus, directrix and axis of symmetry of the parabola are $\left(-5,2\right),\left(-5,-2\right),y=6$ and $\left(x=-5\right)$ respectively.

Explanation of Solution

Given information: The equation of the parabola is $x^2+10x+16y-7=0$ .

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

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

  $\begin{array}{l}{x}^2+10x+16y-7=0\\ x^2+10x=-16y+7\\ {\left(x+5\right)}^2-25=-16y+7\\ {\left(x+5\right)}^2=-16y+32\\ {\left(x+5\right)}^2=-16\left(y-2\right)\\ y-2=-\frac{1}{16}{\left(x+5\right)}^2\end{array}$

Now, compare $y-2=-\frac{1}{16}{\left(x+5\right)}^2$ to equation $y-k=\frac{1}{4c}{\left(x-h\right)}^2$ , we get,

  $\begin{array}{c}h=-5\\ k=2\\ \begin{array}{l}x=h\\ \therefore x=-5\end{array}\\ \begin{array}{l}\frac{1}{4c}=-\frac{1}{16}\\ \therefore c=-4\end{array}\end{array}$

Now,

The vertex are $\left(h,k\right)=\left(-5,2\right)$

The focus is $\left(h,k+c\right)=\left(-5,-2\right)$

The directrix is $y=k-c$

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

The symmetric axis is $x=h=-5$