Chapter 11.1, Problem 19E

To determine

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

Answer to Problem 19E

Focus of parabola is $F\left(0,-\frac{3}{2}\right)$ .

Directrix of parabola is $y=\frac{3}{2}$ .

Focal diameter of parabola Is $6$ .

Explanation of Solution

Focus of parabola is $F\left(0,-\frac{3}{2}\right)$ .

Directrix of parabola is $y=\frac{3}{2}$ .

Focal diameter of parabola Is $6$ .

Given information:

The equation is $x^2+6y=0$ .

First write the equation in a standard form of parabola,

  $\begin{array}{l}{x}^2+6y\\ x^2-\left(-6y\right)\end{array}$

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

  $x^2-4py$

By comparing the equations it can be observed that,

  $\begin{array}{l}4p=-\left(6\right)\\ p=-\frac{3}{2}\text{ [divide]}\end{array}$

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

Since $p=-\frac{3}{2}$ , substitute it,

  $\begin{array}{l}y=-p\\ y=-\left(-\frac{3}{2}\right)\text{ [substitute p by -}\frac{3}{2}]\\ y=\frac{3}{2}\text{ [simplify]}\end{array}$

So the directrix of the parabola is $y=\frac{3}{2}$ .

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

Since $p=-\frac{3}{2}$ , substitute the value of $p$ ,

Thus, the focus of parabola is $F\left(0,-\frac{3}{2}\right)$ .

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

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

Since $p=-\frac{3}{2}$ , substitute the value of $p$ ,

  $\begin{array}{l}\text{|4}p\text{|}\\ \text{|4}·\left(-\frac{3}{2}\right)|\text{ [substitute p by -}\frac{3}{2}]\\ 6\end{array}$

So focal diameter of the parabola is $6$ .

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
-6	-6
-3	-1.5
6	-6
12	-1.5

The graph is obtained as: