Chapter 11.1, Problem 52E

a.

To determine

To find: the equations of the parabolas.

Answer to Problem 52E

The equations of parabolasare $x^2=y,x^2=2y,x^2=4y,x^2=8y$ .

Explanation of Solution

The equations of parabolasare $x^2=y,x^2=2y,x^2=4y,x^2=8y$ .

Given information:

Focus on the positive y-axis.

Focal diameter = $1,2,4,8$

Calculation: it is given that the Focus of parabolas on the positive y-axis

Sothe parabola has vertical axis and $p>0$ .

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

  $x^2-4py$

The focal diameter of parabolas $\left|4p\right|$ .

Since focal diameter of parabola = $1$

So,

  $\begin{array}{l}\left|4p\right|=1\\ \left(4p\right)=1\text{ [remove the mod, because p>0 so it remove with positive sign]}\\ p=\frac{1}{4}\text{ [divide by 4]}\\ \end{array}$

So the equation of parabola is,

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

Since focal diameter of parabola = $2$

So,

  $\begin{array}{l}\left|4p\right|=2\\ \left(4p\right)=2\text{ [remove the mod, because p>0 so it remove with positive sign]}\\ p=\frac{2}{4}\text{ [divide by 4]}\\ p\text{=}\frac{1}{2}\text{ [simplify]}\\ \end{array}$

So the equation of parabola is,

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

Since focal diameter of parabola = $4$

So,

  $\begin{array}{l}\left|4p\right|=4\\ \left(4p\right)=4\text{ [remove the mod, because p>0 so it remove with positive sign]}\\ p=\frac{4}{4}\text{ [divide by 4]}\\ p\text{=}1\text{ [simplify]}\\ \end{array}$

So the equation of parabola is,

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

Since focal diameter of parabola = $8$

So,

  $\begin{array}{l}\left|4p\right|=8\\ \left(4p\right)=8\text{ [remove the mod, because p>0 so it remove with positive sign]}\\ p=\frac{8}{4}\text{ [divide by 4]}\\ p\text{=}2\text{ [simplify]}\\ \end{array}$

So the equation of parabola is,

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

Hence, the equation of parabola is $x^2=y,x^2=2y,x^2=4y,x^2=8y$ .

b.

To determine

To graph the parabolas using graphing device.

Explanation of Solution

Given information: Use graphing device to graph the parabola $x^2=y,x^2=2y,x^2=4y,x^2=8y$ .

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

  $\begin{array}{l}{x}^2=y\\ x^2=2y\\ y_2=\frac{x^2}{2}\text{ [divide by 2]}\\ x^2=4y\\ y_3=\frac{x^2}{4}\text{ [divide by 4]}\\ x^2=8y\\ y_4=\frac{x^2}{8}\text{ [divide by 8]}\end{array}$

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

  $y_4=\frac{x^2}{8},y_3=\frac{x^2}{4},y_2=\frac{x^2}{2},y_1=\frac{x^2}{1}$

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 $y_4=\frac{x^2}{8},y_3=\frac{x^2}{4},y_2=\frac{x^2}{2},y_1=\frac{x^2}{1}$ 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.