The equation of an ellipse is given below. ((x-3)^(2))/(25)+((y+1)^(2))/(169)=1 What are the foci of this ellipse? Choose 1 answer: (A) (3,-13) and (3,11) (B) (9,1) and (-15,1) (c) (-12,-1) and (15,-1) (D) (-3,13) and (-3,-11)

Identify center and axes lengths: Identify the center and lengths of the semi-major and semi-minor axes. The standard form of an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) is the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis. For the given ellipse, h = 3, k = -1, a^2 = 25, and b^2 = 169. Therefore, a = 5 and b = 13.Determine major axis: Determine which axis is the major axis. Since b > a, the major axis is along the y-axis. This means that the foci will be vertically aligned with the center of the ellipse.Calculate distance to foci: Calculate the distance c from the center to each focus. The distance c is found using the equation c^2 = b^2 - a^2. Plugging in the values, we get c^2 = 169 - 25 = 144. Taking the square root gives us c = 12.Find foci coordinates: Find the coordinates of the foci. The foci are located at (h, k ± c) since the major axis is vertical. Substituting the values, we get the foci at (3, -1 ± 12). This gives us the two points (3, -1 + 12) and (3, -1 - 12), which simplifies to (3, 11) and (3, -13).

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The equation of an ellipse is given below.

((x-3)^(2))/(25)+((y+1)^(2))/(169)=1
What are the foci of this ellipse?
Choose 1 answer:
(A)
(3,-13) and
(3,11)
(B)
(9,1) and
(-15,1)
(c)
(-12,-1) and
(15,-1)
(D)
(-3,13) and
(-3,-11)

The equation of an ellipse is given below. $\newline$ $\frac{(x-3)^{2}}{25}+\frac{(y+1)^{2}}{169}=1$ $\newline$ What are the foci of this ellipse? $\newline$ Choose $1$ answer: $\newline$ (A) $(3,-13)$ and $(3,11)$ $\newline$ (B) $(9,1)$ and $(-15,1)$ $\newline$ (C) $(-12,-1)$ and $(15,-1)$ $\newline$ (D) $(-3,13)$ and $(-3,-11)$

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Algebra 2

Find properties of ellipses from equations in general form

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Q. The equation of an ellipse is given below.

\newline

\frac{(x-3)^{2}}{25}+\frac{(y+1)^{2}}{169}=1

\newline

What are the foci of this ellipse?

\newline

Choose

1

answer:

\newline

(A)

(3,-13)

and

(3,11)

\newline

(B)

(9,1)

and

(-15,1)

\newline

(C)

(-12,-1)

and

(15,-1)

\newline

(D)

(-3,13)

and

(-3,-11)

Identify center and axes lengths: Identify the center and lengths of the semi-major and semi-minor axes. $\newline$ The standard form of an ellipse is $(x-h)^2/a^2 + (y-k)^2/b^2 = 1$ , where $(h, k)$ is the center of the ellipse, $a$ is the length of the semi-major axis, and $b$ is the length of the semi-minor axis. For the given ellipse, $h = 3$ , $k = -1$ , $a^2 = 25$ , and $b^2 = 169$ . Therefore, $a = 5$ and $b = 13$ .
Determine major axis: Determine which axis is the major axis. $\newline$ Since b > a, the major axis is along the y-axis. This means that the foci will be vertically aligned with the center of the ellipse.
Calculate distance to foci: Calculate the distance $c$ from the center to each focus. The distance $c$ is found using the equation $c^2 = b^2 - a^2$ . Plugging in the values, we get $c^2 = 169 - 25 = 144$ . Taking the square root gives us $c = 12$ .
Find foci coordinates: Find the coordinates of the foci. $\newline$ The foci are located at $(h, k \pm c)$ since the major axis is vertical. Substituting the values, we get the foci at $(3, -1 \pm 12)$ . This gives us the two points $(3, -1 + 12)$ and $(3, -1 - 12)$ , which simplifies to $(3, 11)$ and $(3, -13)$ .