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The equation of an ellipse is given below.

((x+9)^(2))/(100)+(y^(2))/(64)=1
What are the foci of this ellipse?
Choose 1 answer:
(A)
(-9,-3) and
(-9,3)
(B)
(-9,-6) and
(-9,6)
(C)
(-15,0) and
(-3,0)
(D)
(-12,0) and
(-6,0)

The equation of an ellipse is given below. $\newline$ $\frac{(x+9)^{2}}{100}+\frac{y^{2}}{64}=1$ $\newline$ What are the foci of this ellipse? $\newline$ Choose $1$ answer: $\newline$ (A) $(-9,-3)$ and $(-9,3)$ $\newline$ (B) $(-9,-6)$ and $(-9,6)$ $\newline$ (C) $(-15,0)$ and $(-3,0)$ $\newline$ (D) $(-12,0)$ and $(-6,0)$

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Find properties of ellipses from equations in general form

Full solution

Q. The equation of an ellipse is given below.

\newline

\frac{(x+9)^{2}}{100}+\frac{y^{2}}{64}=1

\newline

What are the foci of this ellipse?

\newline

Choose

1

answer:

\newline

(A)

(-9,-3)

and

(-9,3)

\newline

(B)

(-9,-6)

and

(-9,6)

\newline

(C)

(-15,0)

and

(-3,0)

\newline

(D)

(-12,0)

and

(-6,0)

Identify center and lengths: Identify the center and lengths of the major and minor axes. $\newline$ The given equation of the ellipse is $\frac{(x+9)^{2}}{100}+\frac{y^{2}}{64}=1$ . This is in the standard form of an ellipse equation, which is $\frac{(x-h)^{2}}{a^2}+\frac{(y-k)^{2}}{b^2}=1$ , where $(h, k)$ is the center of the ellipse, $a$ is the semi-major axis length, and $b$ is the semi-minor axis length. For our ellipse, $h = -9$ , $k = 0$ , $a^2 = 100$ , and $b^2 = 64$ . Therefore, $a = 10$ and $\frac{(x-h)^{2}}{a^2}+\frac{(y-k)^{2}}{b^2}=1$ $0$ .
Determine major axis: Determine which axis is the major axis. $\newline$ Since $a^2 > b^2$ , the major axis is along the x-axis, and the minor axis is along the y-axis. This means that the foci will be located along the x-axis, at a distance of $c$ from the center, where $c$ is found using the formula $c = \sqrt{a^2 - b^2}$ .
Calculate distance to foci: Calculate the distance $c$ from the center to each focus. Using the formula $c = \sqrt{a^2 - b^2}$ , we find $c = \sqrt{100 - 64} = \sqrt{36} = 6$ . This means the foci are located $6$ units to the left and right of the center along the x-axis.
Find coordinates of foci: Find the coordinates of the foci. The center of the ellipse is at $(-9, 0)$ . The foci are located at $(-9 - c, 0)$ and $(-9 + c, 0)$ . Substituting the value of $c$ , we get the foci at $(-9 - 6, 0)$ and $(-9 + 6, 0)$ , which simplifies to $(-15, 0)$ and $(-3, 0)$ .

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