Chapter 8.1, Problem 36E

Solution

To sketch: the graph of the parabola ${\left(x-5\right)}^2=20\left(y+2\right)$ by hand.

Given:

The equation of a parabola: ${\left(x-5\right)}^2=20\left(y+2\right)$ .

Graph:

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

Observe that the equation ${\left(x-5\right)}^2=20\left(y+2\right)$ is in the form ${\left(x-h\right)}^2=4p\left(y-k\right)$ where $h=5$ , $k=-2$ and $p=5$ .

Thus, ${\left(x-5\right)}^2=20\left(y+2\right)$ is the equation of the parabola having the vertex $\left(5,-2\right)$ and focus $\left(5,-2+5\right)=\left(5,3\right)$ .

Plot the vertex $\left(5,-2\right)$ and the focus $\left(5,3\right)$ of the parabola on a coordinate grid as shown below:

  

Since the parabola has the vertex $\left(5,-2\right)$ and the focus $\left(5,3\right)$ , it is opened to the upper side of the grid.

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

For that purpose, substitute $y=3$ in the equation ${\left(x-5\right)}^2=20\left(y+2\right)$ and solve it for $x$ :

  $\begin{array}{c}{\left(x-5\right)}^2=20\left(3+2\right)\\ {\left(x-5\right)}^2=100\\ x-5=\pm \sqrt{100}\\ x-5=10,-10\\ x=10+5,-10+5\\ x=15,-5\end{array}$ .

Thus, the points $\left(15,3\right)$ and $\left(-5,3\right)$ lie on the parabola ${\left(x-5\right)}^2=20\left(y+2\right)$ .

Similarly, it may be observed that the points $\left(-15,18\right)$ and $\left(25,18\right)$ lie on the parabola ${\left(x-5\right)}^2=20\left(y+2\right)$ .

Plot the points $\left(15,3\right)$ , $\left(-5,3\right)$ , $\left(-15,18\right)$ and $\left(25,18\right)$ 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: