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Write an equation for an ellipse centered at the origin, which has foci at
(+-12,0) and vertices at
(+-13,0).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm 12,0)$ and vertices at $( \pm 13,0)$ .

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Write equations of ellipses in standard form using properties

Full solution

Q. Write an equation for an ellipse centered at the origin, which has foci at

( \pm 12,0)

and vertices at

( \pm 13,0)

.

Given Foci and Vertices: We are given the foci and vertices of an ellipse centered at the origin. The distance from the center to a focus is denoted by $c$ , and the distance from the center to a vertex is denoted by $a$ . Since the ellipse is centered at the origin, we can directly read these values from the coordinates of the foci and vertices. $\newline$ Foci: $(\pm 12, 0)$ implies $c = 12$ $\newline$ Vertices: $(\pm 13, 0)$ implies $a = 13$
Calculating b: The relationship between $a$ , $b$ , and $c$ for an ellipse is given by the equation $c^2 = a^2 - b^2$ , where $b$ is the distance from the center to the co-vertex. We need to find the value of $b$ using the values of $a$ and $c$ .
Calculate $b$ using the relationship $c^2 = a^2 - b^2$ .
$b$ $0$
$b$ $1$
$b$ $2$
$b$ $3$
Finding the Standard Form Equation: Now that we have $b^2$ , we can find $b$ by taking the square root of $b^2$ . $\newline$ $b = \sqrt{b^2}$ $\newline$ $b = \sqrt{25}$ $\newline$ $b = 5$
Finding the Standard Form Equation: Now that we have $b^2$ , we can find $b$ by taking the square root of $b^2$ . $\newline$ $b = \sqrt{b^2}$ $\newline$ $b = \sqrt{25}$ $\newline$ $b = 5$ With the values of $a$ and $b$ , we can write the standard form equation of the ellipse. Since the ellipse is horizontal (the vertices are on the x-axis), the standard form of the equation is $\left(\frac{x^2}{a^2}\right) + \left(\frac{y^2}{b^2}\right) = 1$ . $\newline$ Substitute $a = 13$ and $b = 5$ into the equation. $\newline$ $\left(\frac{x^2}{13^2}\right) + \left(\frac{y^2}{5^2}\right) = 1$ $\newline$ $\left(\frac{x^2}{169}\right) + \left(\frac{y^2}{25}\right) = 1$

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