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Q 60.

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. Vertices at $\left(2,5\right)$ and $\left(2,-1\right)$:c=2

The equation of the ellipse is $\frac{{\left(x-2\right)}^{2}}{5}+\frac{{\left(y-2\right)}^{2}}{9}=1$and graph of the ellipse is

See the step by step solution

## Step 1. Given information.

Vertices at $\left(2,5\right)$ and $\left(2,-1\right)$: $c=2$

The vertices lies on the line $x=2$, thus the major axis is parallel to the y-axis.

The center is at the mid point of the vertices $\left(2,2\right)$.

The distance of the center from one of the foci is $c=2$.

The distance of the center from one of the vertices is $a=3$.

## Step 2. The equation of ellipse.

Substitute $a=3$ and $c=2$ in ${b}^{2}={a}^{2}-{c}^{2}$ we get.

${b}^{2}={a}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}{b}^{2}=9-4\phantom{\rule{0ex}{0ex}}{b}^{2}=5$

Substituting the values of ${a}^{2}$ and ${b}^{2}$ in the equation of the ellipse we get.

$\frac{{\left(x-2\right)}^{2}}{5}+\frac{{\left(y-2\right)}^{2}}{9}=1$