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Q 63.

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 $\left(1,2\right)$: vertex at $\left(4,2\right)$:contains the point$\left(1,5\right)$

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

See the step by step solution

Step 1 . Given information.

Center at $\left(1,2\right)$: vertex at $\left(4,2\right)$:contains the point $\left(1,5\right)$.

On comparing with the standard equation of the ellipse $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$

Here,$x=1,y=5,h=1,k=2$ and $h+a=4\phantom{\rule{0ex}{0ex}}a=4-1\phantom{\rule{0ex}{0ex}}a=3$

Now put these values in the equation of the ellipse.

$\frac{{\left(1-1\right)}^{2}}{{3}^{2}}+\frac{{\left(5-2\right)}^{2}}{{b}^{2}}=1\phantom{\rule{0ex}{0ex}}\frac{0}{9}+\frac{9}{{b}^{2}}=1\phantom{\rule{0ex}{0ex}}\frac{9}{{b}^{2}}=1\phantom{\rule{0ex}{0ex}}{b}^{2}=9$

Step 2. The equation of ellipse.

Substitute the value of ${a}^{2}=9$ and ${b}^{2}=9$ we get

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

Step 3 .Graph of the ellipse .

The major axis is parallel to the x-axis.So the vertices are $a=3$units left and right of the center $\left(1,2\right)$.Therefore the vertices are localid="1646732539408" ${v}_{1}=\left(-2,2\right)$ and localid="1646732554304" ${v}_{2}=\left(4,2\right)$

Since,

localid="1646732713719" ${c}^{2}={a}^{2}-{b}^{2}\phantom{\rule{0ex}{0ex}}{c}^{2}=9-9\phantom{\rule{0ex}{0ex}}{c}^{2}=0$

Major axis is parallel to the x-axis, foci are localid="1646732722940" $\left(1-0,2\right)$ and localid="1646732737337" $\left(1,2\right)$.

We use the value localid="1646732752572" $b=3$ to find the two points above and below the center as localid="1646732764537" $\left(1,-1\right)$ and localid="1646732779266" $\left(1,5\right)$