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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.Vertex is at $\left(4,-2\right)$ and focus is at $\left(6,-2\right)$.

The equation of a parabola is

${\left(y+2\right)}^{2}=8\left(x-4\right)$.

The points are $\left(6,2\right);\left(6,-6\right)$.

The graph of an equation is

See the step by step solution

## Step 1. Given Information.

The given vertex is at $\left(4,-2\right)$ and focus is at $\left(6,-2\right)$.

## Step 2. Equation of a parabola.

The vertex is at $\left(4,-2\right)$ and focus $\left(6,-2\right)$ both lie on the horizontal line $y=-2$ (the axis of symmetry). The distance a from the vertex to the focus is $a=-2$.

The parabola opens to the right. The form of a equation is

${\left(y-k\right)}^{2}=-4a\left(x-h\right)$.

where $\left(h,k\right)=\left(4,-2\right)$ and $a=-2$. Therefore, the equation is

${\left(y+2\right)}^{2}=8\left(x-4\right)$.

## Step 3. Latus rectum.

The two points that determines the latus rectum by letting $x=6$, so that

${\left(y+2\right)}^{2}=8\left(x-4\right)\phantom{\rule{0ex}{0ex}}{\left(y+2\right)}^{2}=8\left(2\right)\phantom{\rule{0ex}{0ex}}{\left(y+2\right)}^{2}=16\phantom{\rule{0ex}{0ex}}y+2=±4\phantom{\rule{0ex}{0ex}}y+2=4,y+2=-4\phantom{\rule{0ex}{0ex}}y=2,-6.$

The points are $\left(6,2\right)$ and $\left(6,-6\right)$.

## Step 4. Graphing Utility.

The graph of a parabola is