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Q 62.

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. Center at $\left(1,2\right)$: focus at $\left(1,4\right)$: contains the point $\left(2,2\right)$

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

See the step by step solution

## Step 1 . Given information.

Center at $\left(1,2\right)$: focus at $\left(1,4\right)$ contains the point$\left(2,2\right)$.

Since the center and the focus lie on the line $x=1$, thus the major axis is parallel to the y-axis.

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

Since, the point $\left(2,2\right)$ lies on the ellipse

$\frac{{\left(2-1\right)}^{2}}{{b}^{2}}+\frac{{\left(2-2\right)}^{2}}{{a}^{2}}=1\phantom{\rule{0ex}{0ex}}{b}^{2}=1$

## Step 2. The equation of ellipse.

Substitute $c=2$ and $a=1$ we get.

${b}^{2}={a}^{2}-{c}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}={b}^{2}+{c}^{2}\phantom{\rule{0ex}{0ex}}{a}^{2}=1+4\phantom{\rule{0ex}{0ex}}{a}^{2}=5$

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