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Q 59.

Expert-verified
Found in: Page 655

### Precalculus Enhanced with Graphing Utilities

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

# Find an equation for each ellipse. Graph the equation by hand. Foci at $\left(5,1\right)$ and $\left(-1,1\right)$: length of the major axis is $8$.

The equation of the ellipse is $\frac{{\left(x-2\right)}^{2}}{16}+\frac{{\left(y-1\right)}^{2}}{7}=1$ and graph of the ellipse is

See the step by step solution

## Step 1. Given information.

Foci at $\left(5,1\right)$ and $\left(-1,1\right)$:length of the major axis is $8$.

The foci lie on the line$y=1$, so the major axis is parallel to the x-axis.

The two foci $\left(5,1\right)$ and $\left(-1,1\right)$, we can find the center of the ellipse which is the midpoint of these two points.

Thus, the center of the ellipse is $\left(2,1\right)$.

## Step 2. The equation of ellipse.

The length of the major axis $2a$ is given as $8$. So, we have

$2a=8\phantom{\rule{0ex}{0ex}}a=4$

The distance from the center of the ellipse $\left(2,1\right)$ to the focus $\left(5,1\right)$ is $c=3$

Substitute $c=3$ and $a=4$ in

role="math" localid="1646719438001" ${b}^{2}={a}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}{b}^{2}={4}^{2}-{3}^{2}\phantom{\rule{0ex}{0ex}}{b}^{2}=16-9\phantom{\rule{0ex}{0ex}}{b}^{2}=7$

The equation of ellipse is $\frac{{\left(x-2\right)}^{2}}{16}+\frac{{\left(y-1\right)}^{2}}{7}=1$

## Step 3 .Graph of the ellipse

The major axis parallel to the x- axis. So the vertices of the ellipse are $a=4$units left and right of the center $\left(2,1\right)$.

Thus, the vertices are ${v}_{1}=\left(-2,1\right)$ and ${v}_{2}=\left(6,1\right)$.

We use $b=\sqrt{7}$ to find the two points above and below the center. The two points are $\left(2,1+\sqrt{7}\right)$and $\left(2,1-\sqrt{7}\right)$.

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