Write an equation for an ellipse centered at the origin, which has foci at (0,+-sqrt14) and co-vertices at (+-sqrt2,0).

Ellipse Orientation: We have: Foci: (0, ±√14) Co-vertices: (±√2, 0) Choose the orientation of the ellipse. Since the foci are along the y-axis, the ellipse is vertical.Identifying c: We have: Center (h, k): (0, 0) Foci: (0, ±√14) Identify the value of c. c = √14Identifying b: We have: Center (h, k): (0, 0) Co-vertices: (±√2, 0) Identify the value of b. b = √2Calculating a^2: We know: c^2 = a^2 - b^2 We have c = √14 and b = √2. Calculate the value of a^2. a^2 = c^2 + b^2 a^2 = (√14)^2 + (√2)^2 a^2 = 14 + 2 a^2 = 16Finding a: Find the value of a. a = √a^2 a = √16 a = 4Standard Form of the Ellipse: We know: (h, k) = (0, 0) a = 4 b = √2 What would be the standard form of the ellipse? Since the ellipse is vertical, the a^2 term will be under the y^2 term. (x - h)^2/b^2 + (y - k)^2/a^2 = 1 (x - 0)^2/(√2)^2 + (y - 0)^2/4^2 = 1 x^2/2 + y^2/16 = 1

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Write an equation for an ellipse centered at the origin, which has foci at
(0,+-sqrt14) and co-vertices at
(+-sqrt2,0).

Write an equation for an ellipse centered at the origin, which has foci at $(0, \pm \sqrt{14})$ and co-vertices at $( \pm \sqrt{2}, 0)$ .

View step-by-step help

Home

Math Problems

Algebra 2

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 \sqrt{14})

and co-vertices at

( \pm \sqrt{2}, 0)

Ellipse Orientation: We have: $\newline$ Foci: $(0, \pm\sqrt{14})$ $\newline$ Co-vertices: $(\pm\sqrt{2}, 0)$ $\newline$ Choose the orientation of the ellipse. $\newline$ Since the foci are along the y-axis, the ellipse is vertical.
Identifying $c$ : We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Foci: $(0, \pm\sqrt{14})$ $\newline$ Identify the value of $c$ . $\newline$ $c = \sqrt{14}$
Identifying $b$ : We have: $\newline$ Center $(h, k)$ : $(0, 0)$ $\newline$ Co-vertices: $(\pm\sqrt{2}, 0)$ $\newline$ Identify the value of $b$ . $\newline$ $b = \sqrt{2}$
Calculating $a^2$ : We know: $\newline$ $c^2 = a^2 - b^2$ $\newline$ We have $c = \sqrt{14}$ and $b = \sqrt{2}$ . $\newline$ Calculate the value of $a^2$ . $\newline$ $a^2 = c^2 + b^2$ $\newline$ $a^2 = (\sqrt{14})^2 + (\sqrt{2})^2$ $\newline$ $a^2 = 14 + 2$ $\newline$ $a^2 = 16$
Finding $a$ : Find the value of $a$ .
$a = \sqrt{a^2}$
$a = \sqrt{16}$
$a = 4$
Standard Form of the Ellipse: We know: $\newline$ $(h, k) = (0, 0)$ $\newline$ $a = 4$ $\newline$ $b = \sqrt{2}$ $\newline$ What would be the standard form of the ellipse? $\newline$ Since the ellipse is vertical, the $a^2$ term will be under the $y^2$ term. $\newline$ $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$ $\newline$ $\frac{(x - 0)^2}{(\sqrt{2})^2} + \frac{(y - 0)^2}{4^2} = 1$ $\newline$ $\frac{x^2}{2} + \frac{y^2}{16} = 1$