Chapter 9.3, Problem 23WE

To determine

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

Answer to Problem 23WE

The required vertex, focus, directrix and axis of symmetry of the parabola are $\left(-\frac{3}{2},3\right),\left(\frac{1}{2},3\right),x=-\frac{7}{2}$ and $\left(y=3\right)$ respectively.

Explanation of Solution

Given information: The equation of the parabola is $y^2-8x-6y-3=0$ .

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

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

  $\begin{array}{l}{y}^2-8x-6y-3=0\\ y^2-6y=8x+3\\ y^2-6y+9=8x+3+9\\ {\left(y-3\right)}^2=8\left(x-\left(-\frac{3}{2}\right)\right)\\ \left(x-\left(-\frac{3}{2}\right)\right)=\frac{1}{8}{\left(y-3\right)}^2\end{array}$

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

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

Now,

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

The focus is $\left(h+c,k\right)=\left(\frac{-3}{2}+2,3\right)==\left(\frac{1}{2},3\right)$

The directrix is $x=h-c$

  $\begin{array}{l}=-\frac{3}{2}-2\\ \therefore x=-\frac{7}{2}\end{array}$

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

Now, graph the given parabola,