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Q 26.

Expert-verified
Found in: Page 643

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand. line of directix $x=-\frac{1}{2}$ and vertex at the point $\left(0,0\right)$.

The equation of a parabola is ${y}^{2}=2x$. The two points are $\left(\frac{1}{2},1\right)$ and $\left(\frac{1}{2},-1\right)$.

The graph of an equation is

See the step by step solution

## Step 1. Given Information.

The given vertex is at the point $\left(0,0\right)$ and the line of directix is $x=-\frac{1}{2}$.

## Step 2. Equation of a Parabola.

A parabola whose focus is at $\left(\frac{1}{2},0\right)$ and whose directix is the line $x=-\frac{1}{2}$ has its vertex at $\left(0,0\right)$.The equation of a parabola is of the form as

${y}^{2}=4ax\phantom{\rule{0ex}{0ex}}{y}^{2}=2x$

where $a=\frac{1}{2}$.

## Step 3. Latus Rectum.

The two points that determines the latus rectum by letting $x=\frac{1}{2}$.Then,

${y}^{2}=2x\phantom{\rule{0ex}{0ex}}{y}^{2}=2\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}{y}^{2}=1\phantom{\rule{0ex}{0ex}}y=±1$.

The points are $\left(\frac{1}{2},1\right)$ and $\left(\frac{1}{2},-1\right)$.

## Step 4. Graphing Utility.

The graph of a parabola is