Chapter 11.1, Problem 22E

To determine

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

Answer to Problem 22E

Focus of parabola is $F\left(0,-\frac{3}{8}\right)$ .

Directrix of parabola is $y=\frac{3}{8}$ .

Focal diameter of parabola Is $\frac{3}{2}$ .

Explanation of Solution

Focus of parabola is $F\left(0,-\frac{3}{8}\right)$ .

Directrix of parabola is $y=\frac{3}{8}$ .

Focal diameter of parabola Is $\frac{3}{2}$ .

Given information:

The equation is $8x^2+12y=0$ .

First write the equation in a standard form of parabola,

  $\begin{array}{l}8x^2+12y=0\\ x^2=-\frac{12y}{8}\text{ [divide by 8]}\\ x^2-\left(-\frac{3}{2}y\right)\text{ [simplify] }\end{array}$

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

  $x^2-4py$

By comparing the equations it can be observed that,

  $\begin{array}{l}4p=-\left(\frac{3}{2}\right)\\ p=-\frac{3}{8}\text{ [divide]}\end{array}$

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

Since $p=-\frac{3}{8}$ , substitute it,

  $\begin{array}{l}y=-p\\ y=-\left(-\frac{3}{8}\right)\text{ [substitute p by -}\frac{3}{8}]\\ y=\frac{3}{8}\text{ [simplify]}\end{array}$

So the directrix of the parabola is $y=\frac{3}{8}$ .

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

Since $p=-\frac{3}{8}$ , substitute the value of $p$ ,

Thus, the focus of parabola is $F\left(0,-\frac{3}{8}\right)$ .

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

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

Since $p=-\frac{3}{8}$ , substitute the value of $p$ ,

  $\begin{array}{l}\text{|4}p\text{|}\\ \text{|4}·\left(-\frac{3}{8}\right)|\text{ [substitute p by -}\frac{3}{8}]\\ \frac{3}{2}\end{array}$

So focal diameter of the parabola is $\frac{3}{2}$ .

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

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

To sketch the graph,

x	y
1	-0.66
2	-2.66
-2	-2.66
1	-0.66

The graph is obtained as: