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Q 59.

Expert-verifiedFound in: Page 655

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Find an equation for each ellipse. Graph the equation by hand.

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

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

Foci at $(5,1)$ and $(-1,1)$: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 $(5,1)$ and $(-1,1)$, we can find the center of the ellipse which is the midpoint of these two points.

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

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 $(2,1)$ to the focus $(5,1)$ 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 $\frac{{(x-2)}^{2}}{16}+\frac{{(y-1)}^{2}}{7}=1$

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

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

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

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