The equation of an ellipse is given below. $\newline$ $\frac{x^{2}}{15}+\frac{(y-5)^{2}}{20}=1$ $\newline$ What are the foci of this ellipse? $\newline$ Choose $1$ answer: $\newline$ (A) $(0,10)$ and $(0,0)$ $\newline$ (B) $(0,5+\sqrt{5})$ and $(0,5-\sqrt{5})$ $\newline$ (C) $(5,5)$ and $(-5,5)$ $\newline$ (D) $(0+\sqrt{5}, 5)$ and $(0-\sqrt{5}, 5)$

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Algebra 2

Write equations of ellipses in standard form using properties

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Q. The equation of an ellipse is given below.

\newline

\frac{x^{2}}{15}+\frac{(y-5)^{2}}{20}=1

\newline

What are the foci of this ellipse?

\newline

Choose

1

answer:

\newline

(A)

(0,10)

and

(0,0)

\newline

(B)

(0,5+\sqrt{5})

and

(0,5-\sqrt{5})

\newline

(C)

(5,5)

and

(-5,5)

\newline

(D)

(0+\sqrt{5}, 5)

and

(0-\sqrt{5}, 5)

Identify values of $a^2$ and $b^2$ : Identify the values of $a^2$ and $b^2$ from the given standard form of the ellipse equation. $\newline$ 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 semi-major axis, and $b$ is the semi-minor axis. $\newline$ From the given equation $(x^{2})/(15)+((y-5)^{2})/(20)=1$ , we can see that $a^2 = 20$ and $b^2$ $0$ .
Determine major axis: Determine which axis is the major axis. $\newline$ Since $a^2 > b^2$ , the major axis is along the y-direction. This means that the foci will be located along the y-axis at a distance of $c$ from the center, where $c$ is the distance from the center to a focus.
Calculate value of c: Calculate the value of $c$ using the relationship $c^2 = a^2 - b^2$ . $c^2 = 20 - 15$ $c^2 = 5$ $c = \sqrt{5}$
Identify center of the ellipse: Identify the center of the ellipse. $\newline$ The center of the ellipse is at $(h,k)$ , which can be determined from the given equation. Since there is no $(x-h)^2$ term, $h=0$ . The $(y-k)^2$ term is $(y-5)^2$ , so $k=5$ . Therefore, the center of the ellipse is at $(0,5)$ .
Determine coordinates of the foci: Determine the coordinates of the foci. $\newline$ The foci are located at a distance of $c$ from the center along the major axis. Since the major axis is vertical, the foci will have the same $x$ -coordinate as the center, which is $0$ , and $y$ -coordinates will be $k \pm c$ . $\newline$ So the coordinates of the foci are $(0, 5 + \sqrt{5})$ and $(0, 5 - \sqrt{5})$ .