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Write an equation for an ellipse centered at the origin, which has foci at
(+-8,0) and vertices at
(+-17,0).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm 8,0)$ and vertices at $( \pm 17,0)$ .

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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 8,0)

and vertices at

( \pm 17,0)

.

Identify Orientation: We have: $\newline$ Foci: $(\pm 8, 0)$ $\newline$ Vertices: $(\pm 17, 0)$ $\newline$ Choose the orientation of the ellipse. $\newline$ Since the foci and vertices are on the $x$ -axis, the ellipse is horizontal.
Find Value of $a$ : We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Vertices: $(\pm 17, 0)$ $\newline$ Identify the value of $a$ (the distance from the center to a vertex). $\newline$ $a = |\pm 17 - 0|$ $\newline$ = $17$
Find Value of c: We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Foci: $(\pm8, 0)$ $\newline$ Identify the value of $c$ (the distance from the center to a focus). $\newline$ $c = |\pm8 - 0|$ $\newline$ = $8$
Use Relationship to Find $b$ : We know: $\newline$ $a = 17$ $\newline$ $c = 8$ $\newline$ Use the relationship $c^2 = a^2 - b^2$ to find $b$ (the distance from the center to a co-vertex). $\newline$ $c^2 = a^2 - b^2$ $\newline$ $8^2 = 17^2 - b^2$ $\newline$ $64 = 289 - b^2$ $\newline$ $b^2 = 289 - 64$ $\newline$ $b^2 = 225$ $\newline$ $a = 17$ $0$ $\newline$ $a = 17$ $1$
Write Standard Form: We know: $\newline$ $(h, k) = (0, 0)$ $\newline$ $a = 17$ $\newline$ $b = 15$ $\newline$ Write the standard form of the ellipse. $\newline$ $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ $\newline$ $\frac{(x - 0)^2}{17^2} + \frac{(y - 0)^2}{15^2} = 1$ $\newline$ $\frac{x^2}{17^2} + \frac{y^2}{15^2} = 1$ $\newline$ $\frac{x^2}{289} + \frac{y^2}{225} = 1$

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