Chapter 11.1, Problem 20E

To determine

To find: the focus, directrix and focal diameter of the parabola and sketch the graph.

Answer to Problem 20E

Focus of parabola is $F\left(\frac{1}{28},0\right)$ .

Directrix of parabola is $x=-\frac{1}{28}$ .

Focal diameter of parabola Is $\frac{1}{7}$ .

Explanation of Solution

Focus of parabola is $F\left(\frac{1}{28},0\right)$ .

Directrix of parabola is $x=-\frac{1}{28}$ .

Focal diameter of parabola Is $\frac{1}{7}$ .

Given information:

The equation is $x-7y^2=0$ .

First write the equation in a standard form of parabola,

  $\begin{array}{l}x-7y^2=0\\ y^2=\frac{x}{7}\text{ [divide by }\frac{1}{7}\text{]}\\ y^2-\frac{\left(x\right)}{7}\end{array}$

Comparing the standard form of the parabola to the general equation,

  $x^2-4py$

By comparing the equations it can be observed that,

  $\begin{array}{l}4p=\frac{1}{7}\\ p=\frac{1}{28}\text{ [divide]}\end{array}$

Directrix: the directrix of parabola is given by $x=-p$ ,

Since $p=\frac{1}{28}$ , substitute it,

  $\begin{array}{l}x=-p\\ x=-\left(\frac{1}{28}\right)\text{ [substitute p by }\frac{1}{28}]\\ x=-\frac{1}{28}\text{ [simplify]}\end{array}$

So the directrix of the parabola is $x=-\frac{1}{28}$ .

Focus: the focus of parabola is given as $F\left(p,0\right)$ ,

Since $p=\frac{1}{28}$ , substitute the value of $p$ ,

Thus, the focus of parabola is $F\left(\frac{1}{28},0\right)$ .

Vertices : $\left(0,0\right)$

Focal diameter of the parabola: $\text{|4}p\text{|}$

Since $p=\frac{1}{28}$ , substitute the value of $p$ ,

  $\begin{array}{l}\text{|4}p\text{|}\\ \text{|4}·\left(\frac{1}{28}\right)|\text{ [substitute p by }\frac{1}{28}]\\ \frac{1}{7}\end{array}$

So focal diameter of the parabola is $\frac{1}{7}$ .

Since the $\text{p>0}$ , so the parabola is opened to the right,

Use the above information together with some additional values which is show in table below

To sketch the graph,

x	y
7	1, -1
14	1.41, -1.41
21	1.73, -1.73
28	2, -2

The graph is obtained as: