What is the center of the ellipse x^2 + 2y^2 - 12 = 0? Write your answer in simplified, rationalized form. (______,______)

Isolate x and y terms: Now, we need to get the equation in standard form for an ellipse. Divide both sides by 12 to isolate the terms with x and y. x^2/12 + 2y^2/12 = 12/12Simplify the equation: Simplify the equation by reducing the fractions. x^2/12 + y^2/6 = 1Identify the center: The standard form of an ellipse is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) is the center. Our equation is already in this form with h = 0 and k = 0. So, the center of the ellipse is (0, 0).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

What is the center of the ellipse $x^2 + 2y^2 - 12 = 0$ ? $\newline$ Write your answer in simplified, rationalized form. $\newline$ (,)

View step-by-step help

Home

Math Problems

Algebra 2

Find properties of ellipses from equations in general form

Full solution

Q. What is the center of the ellipse

x^2 + 2y^2 - 12 = 0

\newline

Write your answer in simplified, rationalized form.

\newline

(______,______)

Isolate x and y terms: Now, we need to get the equation in standard form for an ellipse. $\newline$ Divide both sides by $12$ to isolate the terms with $x$ and $y$ . $\newline$ $\frac{x^2}{12} + \frac{2y^2}{12} = \frac{12}{12}$
Simplify the equation: Simplify the equation by reducing the fractions. $\frac{x^2}{12} + \frac{y^2}{6} = 1$
Identify the center: The standard form of an ellipse is $(x-h)^2/a^2 + (y-k)^2/b^2 = 1$ , where $(h, k)$ is the center. $\newline$ Our equation is already in this form with $h = 0$ and $k = 0$ . $\newline$ So, the center of the ellipse is $(0, 0)$ .