The equation of an ellipse is given below. (x^(2))/(67)+(y^(2))/(57)=1 What are the foci of this ellipse? Choose 1 answer: (A) (10,0) and (-10,0) (B) (0,10) and (0,-10) (C) (sqrt10,0) and (-sqrt10,0) (D) (0,sqrt10) and (0,-sqrt10)

Identify Ellipse Equation: The given equation of the ellipse is (x^2)/67 + (y^2)/57 = 1. To find the foci, we need to determine the values of a and b, where a is the semi-major axis and b is the semi-minor axis. The larger denominator corresponds to the square of the semi-major axis, a^2, and the smaller denominator corresponds to the square of the semi-minor axis, b^2.Determine Semi-Major and Semi-Minor Axes: In the given equation, a^2 = 67 and b^2 = 57. Therefore, a = √67 and b = √57. Since a^2 > b^2, the ellipse is horizontal, and the foci will be located along the x-axis.Calculate Distance to Foci: To find the foci, we use the formula c^2 = a^2 - b^2, where c is the distance from the center to each focus. Let's calculate c. c^2 = 67 - 57 c^2 = 10 c = √10Locate Foci on x-Axis: Since the ellipse is centered at the origin and is horizontal, the foci will be at (±c, 0). Therefore, the foci are at (±√10, 0).Final Answer Comparison: The correct answer is (C) (√10, 0) and (-√10, 0), which matches one of the given choices.

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The equation of an ellipse is given below.

(x^(2))/(67)+(y^(2))/(57)=1
What are the foci of this ellipse?
Choose 1 answer:
(A)
(10,0) and
(-10,0)
(B)
(0,10) and
(0,-10)
(C)
(sqrt10,0) and
(-sqrt10,0)
(D)
(0,sqrt10) and
(0,-sqrt10)

The equation of an ellipse is given below. $\newline$ $\frac{x^{2}}{67}+\frac{y^{2}}{57}=1$ $\newline$ What are the foci of this ellipse? $\newline$ Choose $1$ answer: $\newline$ (A) $(10,0)$ and $(-10,0)$ $\newline$ (B) $(0,10)$ and $(0,-10)$ $\newline$ (C) $(\sqrt{10}, 0)$ and $(-\sqrt{10}, 0)$ $\newline$ (D) $(0, \sqrt{10})$ and $(0,-\sqrt{10})$

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Algebra 2

Write equations of ellipses in standard form using properties

Full solution

Q. The equation of an ellipse is given below.

\newline

\frac{x^{2}}{67}+\frac{y^{2}}{57}=1

\newline

What are the foci of this ellipse?

\newline

Choose

1

answer:

\newline

(A)

(10,0)

and

(-10,0)

\newline

(B)

(0,10)

and

(0,-10)

\newline

(C)

(\sqrt{10}, 0)

and

(-\sqrt{10}, 0)

\newline

(D)

(0, \sqrt{10})

and

(0,-\sqrt{10})

Identify Ellipse Equation: The given equation of the ellipse is $(\frac{x^2}{67} + \frac{y^2}{57} = 1)$ . To find the foci, we need to determine the values of $a$ and $b$ , where $a$ is the semi-major axis and $b$ is the semi-minor axis. The larger denominator corresponds to the square of the semi-major axis, $a^2$ , and the smaller denominator corresponds to the square of the semi-minor axis, $b^2$ .
Determine Semi-Major and Semi-Minor Axes: In the given equation, $a^2 = 67$ and $b^2 = 57$ . Therefore, $a = \sqrt{67}$ and $b = \sqrt{57}$ . Since a^2 > b^2, the ellipse is horizontal, and the foci will be located along the $x$ -axis.
Calculate Distance to Foci: To find the foci, we use the formula $c^2 = a^2 - b^2$ , where $c$ is the distance from the center to each focus. Let's calculate $c$ . $\newline$ $c^2 = 67 - 57$ $\newline$ $c^2 = 10$ $\newline$ $c = \sqrt{10}$
Locate Foci on x-Axis: Since the ellipse is centered at the origin and is horizontal, the foci will be at $(\pm c, 0)$ . Therefore, the foci are at $(\pm\sqrt{10}, 0)$ .
Final Answer Comparison: The correct answer is (C) $(\sqrt{10}, 0)$ and $(-\sqrt{10}, 0)$ , which matches one of the given choices.