Chapter 8.1, Problem 35E

Solution

To sketch: the graph of the parabola ${\left(y-1\right)}^2=8\left(x+3\right)$ by hand.

Given:

The equation of a parabola: ${\left(y-1\right)}^2=8\left(x+3\right)$ .

Graph:

The general form of the equation of a parabola that has the vertex $\left(h,k\right)$ and focus $\left(h+p,k\right)$ is ${\left(y-k\right)}^2=4p\left(x-h\right)$ .

Observe that the equation ${\left(y-1\right)}^2=8\left(x+3\right)$ is in the form ${\left(y-k\right)}^2=4p\left(x-h\right)$ where $h=-3$ , $k=1$ and $p=2$ .

Thus, ${\left(y-1\right)}^2=8\left(x+3\right)$ is the equation of the parabola having the vertex $\left(-3,1\right)$ and the focus $\left(-3+2,1\right)$ .

Plot the vertex $\left(-3,1\right)$ and the focus $\left(-1,1\right)$ of the parabola on a coordinate grid as shown below:

  

Since the parabola has the vertex $\left(-3,1\right)$ and the focus $\left(-1,1\right)$ , it is opened to the right side of the grid.

Now, find few points on the parabola for the purpose of drawing it accurately.

For that purpose, substitute $x=-1$ in the equation ${\left(y-1\right)}^2=8\left(x+3\right)$ and solve it for $y$ :

  $\begin{array}{l}{\left(y-1\right)}^2=8\left(-1+3\right)\\ {\left(y-1\right)}^2=16\\ y-1=\pm \sqrt{16}\\ y-1=4,-4\\ y=4+1,-4+1\\ y=5,-3\end{array}$ .

Thus, the points $\left(-1,-3\right)$ and $\left(-1,5\right)$ lie on the parabola ${\left(y-1\right)}^2=8\left(x+3\right)$ .

Similarly, it may be observed that the points $\left(5,-7\right)$ and $\left(5,9\right)$ lie on the parabola ${\left(y-1\right)}^2=8\left(x+3\right)$ .

Plot the points $\left(-1,-3\right)$ , $\left(-1,5\right)$ , $\left(5,-7\right)$ and $\left(5,9\right)$ along with the vertex and the focus of the parabola on the same grid:

  

Then, join the points plotted previously and the vertex of the parabola using a smooth curve to obtain the graph of the parabola as shown below: