Chapter 11.1, Problem 8E

To determine

To match: the equation with the graph.

Answer to Problem 8E

  

Explanation of Solution

Given information:

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

First write the equation in a standard form of parabola,

  $\begin{array}{l}2x^2=y\\ x^2=\frac{y}{2}\text{ [divide by 2]}\\ x^2-\frac{y}{2}\end{array}$

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

  $x^2-4py$

By comparing the equations it can be observed that,

  $\begin{array}{l}4p=\frac{1}{2}\\ p=\frac{1}{8}\text{ [divide]}\end{array}$

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

Since $p=\frac{1}{8}$ , substitute it,

  $\begin{array}{l}y=-p\\ y=-\left(\frac{1}{8}\right)\text{ [substitute p by }\frac{1}{8}]\\ y=-\frac{1}{8}\text{ [simplify]}\end{array}$

So the directrix of the parabola is $y=-\frac{1}{8}$ .

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

Since $p=\frac{1}{8}$ , substitute the value of $p$ ,

Thus, the focus of parabola is $F\left(0,\frac{1}{8}\right)$ .

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

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,

y	x
1	0.707
2	1
3	1.22
4	1.41

The graph is obtained as: