Write an equation for an ellipse centered at the origin, which has foci at (+-sqrt3,0) and co-vertices at (0,+-9).

Ellipse Orientation: We have: Foci: (±√3, 0) Co-vertices: (0, ±9) Choose the orientation of the ellipse. Since the foci are on the x-axis, the ellipse is horizontal.Finding c: We have: Center (h, k): (0, 0) Foci: (±√3, 0) Identify the value of c (distance from center to foci). c = √((√3 - 0)^2 + (0 - 0)^2) = √(3)Finding b: We have: Center (h, k): (0, 0) Co-vertices: (0, ±9) Identify the value of b (distance from center to co-vertices). b = √((0 - 0)^2 + (9 - 0)^2) = √(81) = 9Finding a: We know the relationship between a, b, and c for an ellipse is c^2 = a^2 - b^2. We already have c = √3 and b = 9. Now we need to find a. c^2 = a^2 - b^2 (√3)^2 = a^2 - 9^2 3 = a^2 - 81 a^2 = 3 + 81 a^2 = 84 a = √84 a = 2√21Standard Form of the Ellipse: We know: (h, k) = (0, 0) a = 2√21 b = 9 What would be the standard form of the ellipse? (x - 0)^2/(2√21)^2 + (y - 0)^2/9^2 = 1 x^2/(2√21)^2 + y^2/9^2 = 1 x^2/(4*21) + y^2/81 = 1 x^2/84 + y^2/81 = 1

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Write an equation for an ellipse centered at the origin, which has foci at
(+-sqrt3,0) and co-vertices at
(0,+-9).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm \sqrt{3}, 0)$ and co-vertices at $(0, \pm 9)$ .

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Algebra 2

Write equations of ellipses in standard form using properties

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

( \pm \sqrt{3}, 0)

and co-vertices at

(0, \pm 9)

Ellipse Orientation: We have: $\newline$ Foci: $(\pm\sqrt{3}, 0)$ $\newline$ Co-vertices: $(0, \pm9)$ $\newline$ Choose the orientation of the ellipse. $\newline$ Since the foci are on the x-axis, the ellipse is horizontal.
Finding c: We have: $\newline$ Center $(h, k): (0, 0)$ $\newline$ Foci: $(\pm\sqrt{3}, 0)$ $\newline$ Identify the value of $c$ (distance from center to foci). $\newline$ $c = \sqrt{(\sqrt{3} - 0)^2 + (0 - 0)^2}$ $\newline$ $= \sqrt{3}$
Finding $b$ : We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Co-vertices: $(0, \pm 9)$ $\newline$ Identify the value of $b$ (distance from center to co-vertices). $\newline$ $b = \sqrt{((0 - 0)^2 + (9 - 0)^2)}$ $\newline$ $= \sqrt{(81)}$ $\newline$ $= 9$
Finding $a$ : We know the relationship between $a$ , $b$ , and $c$ for an ellipse is $c^2 = a^2 - b^2$ . We already have $c = \sqrt{3}$ and $b = 9$ . Now we need to find $a$ . $c^2 = a^2 - b^2$ $(\sqrt{3})^2 = a^2 - 9^2$ $a$ $0$ $a$ $1$ $a$ $2$ $a$ $3$ $a$ $4$
Standard Form of the Ellipse: We know: $\newline$ $(h, k) = (0, 0)$ $\newline$ $a = 2\sqrt{21}$ $\newline$ $b = 9$ $\newline$ What would be the standard form of the ellipse? $\newline$ $\frac{(x - 0)^2}{(2\sqrt{21})^2} + \frac{(y - 0)^2}{9^2} = 1$ $\newline$ $\frac{x^2}{(2\sqrt{21})^2} + \frac{y^2}{9^2} = 1$ $\newline$ $\frac{x^2}{4\cdot21} + \frac{y^2}{81} = 1$ $\newline$ $\frac{x^2}{84} + \frac{y^2}{81} = 1$