Chapter 8.1, Problem 32E

Solution

To sketch: the graph of the parabola $x^2=8y$ by hand.

Given:

The equation of a parabola: $x^2=8y$ .

Graph:

The general form of the equation of a parabola that has the vertex $\left(0,0\right)$ and focus $\left(0,p\right)$ is $x^2=4py$ .

Observe that the equation $x^2=8y$ is in the form $x^2=4py$ where $p=2$ .

Thus, $x^2=8y$ is the equation of the parabola having the vertex $\left(0,0\right)$ and focus $\left(0,2\right)$ .

Plot the vertex $\left(0,0\right)$ and the focus $\left(0,2\right)$ of the parabola on a coordinate grid as shown below:

  

Since the parabola has the vertex $\left(0,0\right)$ and the focus $\left(0,2\right)$ , it is opened to the upper side of the $y$ -axis.

That is, the possible $y$ values of the parabola $x^2=8y$ are all positive.

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

For that purpose, substitute $y=2$ in the equation $x^2=8y$ and solve it for $x$ :

  $\begin{array}{c}{x}^2=8\cdot 2\\ x^2=16\\ x=\pm \sqrt{16}\\ x=4,-4\end{array}$ .

Thus, the points $\left(4,2\right)$ and $\left(-4,2\right)$ lie on the parabola $x^2=8y$ .

Similarly, it may be observed that the points $\left(8,8\right)$ and $\left(-8,8\right)$ lie on the parabola $x^2=8y$ .

Plot the points $\left(4,2\right)$ , $\left(-4,2\right)$ , $\left(8,8\right)$ and $\left(-8,8\right)$ 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: