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Write an equation for an ellipse centered at the origin, which has foci at
(0,+-3) and co-vertices at
(+-2,0).

Write an equation for an ellipse centered at the origin, which has foci at $(0, \pm 3)$ and co-vertices at $( \pm 2,0)$ .

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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

(0, \pm 3)

and co-vertices at

( \pm 2,0)

.

Ellipse Orientation: We have: $\newline$ Foci: $(0, \pm 3)$ $\newline$ Co-vertices: $(\pm 2, 0)$ $\newline$ Choose the orientation of the ellipse. $\newline$ Since the foci are on the y-axis, the ellipse is vertical.
Finding $c$ : We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Foci: $(0, \pm 3)$ $\newline$ Identify the value of $c$ (distance from center to foci). $\newline$ $c = \sqrt{(0 - 0)^2 + (3 - 0)^2}$ $\newline$ = \sqrt{ $0$ ^ $2$ + $3$ ^ $2$ }\) $\newline$ = \sqrt{ $9$ }\) $\newline$ = $3$
Finding $b$ : We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Co-vertices: $(\pm 2, 0)$ $\newline$ Identify the value of $b$ (distance from center to co-vertices). $\newline$ $b = \sqrt{(2 - 0)^2 + (0 - 0)^2}$ $\newline$ $= \sqrt{2^2 + 0^2}$ $\newline$ $= \sqrt{4}$ $\newline$ $= 2$
Finding $a$ : We know the relationship between $a$ , $b$ , and $c$ for an ellipse is $c^2 = a^2 - b^2$ . We have already found $c = 3$ and $b = 2$ . Now we need to find $a$ . $c^2 = a^2 - b^2$ $3^2 = a^2 - 2^2$ $a$ $0$ $a$ $1$ $a$ $2$ $a$ $3$
Standard Form of the Ellipse: We know: $\newline$ $(h, k) = (0, 0)$ $\newline$ $a = \sqrt{13}$ $\newline$ $b = 2$ $\newline$ What would be the standard form of the ellipse? $\newline$ $\frac{(x - 0)^2}{b^2} + \frac{(y - 0)^2}{a^2} = 1$ $\newline$ $\frac{x^2}{2^2} + \frac{y^2}{13} = 1$ $\newline$ $\frac{x^2}{4} + \frac{y^2}{13} = 1$

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