Chapter 11.1, Problem 11E

To determine

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

Answer to Problem 11E

Focus of parabola is $F\left(0,\frac{9}{4}\right)$ .

Directrix of parabola is $y=-\frac{9}{4}$ .

Focal diameter of parabola Is $9$ .

Explanation of Solution

Focus of parabola is $F\left(0,\frac{9}{4}\right)$ .

Directrix of parabola is $y=-\frac{9}{4}$ .

Focal diameter of parabola Is $9$ .

Given information:

The equation is $x^2=9y$ .

First write the equation in a standard form of parabola,

  $x^2-\left(9y\right)$

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

  $x^2-4py$

By comparing the equations it can be observed that,

  $\begin{array}{l}4p=9\\ p=\frac{9}{4}\text{ [divide]}\end{array}$

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

Since $p=\frac{9}{4}$ , substitute it,

  $\begin{array}{l}y=-p\\ y=-\left(\frac{9}{4}\right)\text{ [substitute p by }\frac{9}{4}]\\ y=-\frac{9}{4}\text{ [simplify]}\end{array}$

So the directrix of the parabola is $y=-\frac{9}{4}$ .

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

Since $p=\frac{9}{4}$ , substitute the value of $p$ ,

Thus, the focus of parabola is $F\left(0,\frac{9}{4}\right)$ .

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

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

Since $p=\frac{9}{4}$ , substitute the value of $p$ ,

  $\begin{array}{l}\text{|4}p\text{|}\\ \text{|4}·\frac{9}{4}|\text{ [substitute p by }\frac{9}{4}]\\ 9\end{array}$

So focal diameter of the parabola is $9$ .

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

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

To sketch the graph,

x	y
-1	0.11
-2	0.44
-3	1
-4	1.77

The graph is obtained as: