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Write an equation for an ellipse centered at the origin, which has foci at
(+-sqrt13,0) and co-vertices at
(0,+-11).

Write an equation for an ellipse centered at the origin, which has foci at $( \pm \sqrt{13}, 0)$ and co-vertices at $(0, \pm 11)$ .

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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 \sqrt{13}, 0)

and co-vertices at

(0, \pm 11)

.

Identify Orientation: We have: $\newline$ Foci: $(\pm\sqrt{13}, 0)$ $\newline$ Co-vertices: $(0, \pm11)$ $\newline$ Choose the orientation of the ellipse. $\newline$ Since the foci are on the x-axis, the ellipse is horizontal.
Identify Center and Foci: We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Foci: $(\pm\sqrt{13}, 0)$ $\newline$ Identify the value of $c$ . $\newline$ $c = \sqrt{13}$
Identify Co-vertices: We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Co-vertices: $(0, \pm 11)$ $\newline$ Identify the value of $b$ . $\newline$ $b = 11$
Calculate Value of a: We know the relationship between $a$ , $b$ , and $c$ in an ellipse is $c^2 = a^2 - b^2$ . We already have $c = \sqrt{13}$ and $b = 11$ . Calculate the value of $a$ . $a^2 = c^2 + b^2$ $a^2 = (\sqrt{13})^2 + 11^2$ $a^2 = 13 + 121$ $b$ $0$ $b$ $1$
Standard Form of Ellipse: We know: $\newline$ $(h, k) = (0, 0)$ $\newline$ $a = \sqrt{134}$ $\newline$ $b = 11$ $\newline$ What would be the standard form of the ellipse? $\newline$ $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ $\newline$ $\frac{(x - 0)^2}{(\sqrt{134})^2} + \frac{(y - 0)^2}{11^2} = 1$ $\newline$ $\frac{x^2}{134} + \frac{y^2}{121} = 1$

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