Chapter 11.1, Problem 51E

a.

To determine

To find: the equations of the parabolas.

Answer to Problem 51E

The equations of parabolasare $x^2=-2y,x^2=-4y,x^2=-16y,x^2=-32y$ .

Explanation of Solution

The equations of parabolasare $x^2=-2y,x^2=-4y,x^2=-16y,x^2=-32y$ .

Given information:

Directrix $y=\frac{1}{2},y=1,y=4\text{ and }y=8$

Calculation: from the given directrix it can be observed that the parabola has vertical axis.

So the equation of parabola should be in the following form,

  $x^2-4py$

For parabola with verticalaxis, directrix is $y=-p$ .

To comparing the directrix it can be observed that,

  $p=-\frac{1}{2},-1,-4,-8$

So the equation of parabola is,

  $\begin{array}{l}{x}^2-4py\\ x^2-4\left(-\frac{1}{2}\right)y\text{ [substitute p by -}\frac{1}{2}\text{]}\\ x^2+2y\text{ [divide]}\\ x^2+2y=0\text{ [simplify]}\\ x^2=-2y\end{array}$

  $\begin{array}{l}{x}^2-4py\\ x^2-4\left(-1\right)y\text{ [substitute p by -}1\text{]}\\ x^2+4y\text{ [multiply]}\\ x^2+4y=0\text{ [simplify]}\\ x^2=-4y\end{array}$

  $\begin{array}{l}{x}^2-4py\\ x^2-4\left(-4\right)y\text{ [substitute p by -4]}\\ x^2+16y\text{ [multiply]}\\ x^2+16y=0\text{ [simplify]}\\ x^2=-16y\end{array}$

  $\begin{array}{l}{x}^2-4py\\ x^2-4\left(-8\right)y\text{ [substitute p by -8]}\\ x^2+32y\text{ [multiply]}\\ x^2+32y=0\text{ [simplify]}\\ x^2=-32y\end{array}$

Hence, the equation of parabola is $x^2=-2y,x^2=-4y,x^2=-16y,x^2=-32y$ .

b.

To determine

To graph the parabolas using graphing device.

Explanation of Solution

Given information: Use graphing device to graph the parabola $x^2=-2y,x^2=-4y,x^2=-16y,x^2=-32y$ .

Graph: to draw the graph using a graphing device, it is needed to solve equation for y.

  $\begin{array}{l}{x}^2=-2y\\ y_1=\frac{x^2}{-2}\text{ [divide by -2]}\\ x^2=-4y\\ y_2=\frac{x^2}{-4}\text{ [divide by -4]}\\ x^2=-16y\\ y_3=\frac{x^2}{-16}\text{ [divide by -16]}\\ x^2=-32y\\ y_4=\frac{x^2}{-32}\text{ [divide by -32]}\end{array}$

To obtain the graph of the parabola, graph the following functions,

  $y_4=\frac{x^2}{-32},y_3=\frac{x^2}{-16},y_2=\frac{x^2}{-4},y_1=\frac{x^2}{-2}$

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $y_4=\frac{x^2}{-32},y_3=\frac{x^2}{-16},y_2=\frac{x^2}{-4},y_1=\frac{x^2}{-2}$ which is required to graph.

Step 4: Press GRAPH button to graph the function.

The graph is obtained as:

  

Interpretation:

From the above graph, it can be observed that when value of p increases the graph of parabolas stretch.