Chapter 11.1, Problem 21E

To determine

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

Answer to Problem 21E

Focus of parabola is $F\left(-\frac{5}{12},0\right)$ .

Directrix of parabola is $x=\frac{5}{12}$ .

Focal diameter of parabola Is $\frac{5}{3}$ .

Explanation of Solution

Focus of parabola is $F\left(-\frac{5}{12},0\right)$ .

Directrix of parabola is $x=\frac{5}{12}$ .

Focal diameter of parabola Is $\frac{5}{3}$ .

Given information:

The equation is $5x+3y^2=0$ .

First write the equation in a standard form of parabola,

  $\begin{array}{l}5x+3y^2=0\\ y^2=\frac{5x}{-3}\text{ [divide by -3]}\\ y^2-\left(-\frac{\left(5x\right)}{3}\right)\end{array}$

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

  $x^2-4py$

By comparing the equations it can be observed that,

  $\begin{array}{l}4p=-\frac{5}{3}\\ p=-\frac{5}{12}\text{ [divide]}\end{array}$

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

Since $p=-\frac{5}{12}$ , substitute it,

  $\begin{array}{l}x=-p\\ x=-\left(-\frac{5}{12}\right)\text{ [substitute p by -}\frac{5}{12}]\\ x=\frac{5}{12}\text{ [simplify]}\end{array}$

So the directrix of the parabola is $x=\frac{5}{12}$ .

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

Since $p=-\frac{5}{12}$ , substitute the value of $p$ ,

Thus, the focus of parabola is $F\left(-\frac{5}{12},0\right)$ .

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

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

Since $p=-\frac{5}{12}$ , substitute the value of $p$ ,

  $\begin{array}{l}\text{|4}p\text{|}\\ \text{|4}·\left(-\frac{5}{12}\right)|\text{ [substitute p by -}\frac{5}{12}]\\ \frac{5}{3}\end{array}$

So focal diameter of the parabola is $\frac{5}{3}$ .

Since the $\text{p<0}$ , so the parabola is opened to the left,

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

To sketch the graph,

x	y
-1	1.28, -1.28
-2	1.82, -1.82
-3	2.23, -2.23
-4	2.58, 2.58

The graph is obtained as: