What is the center of the ellipse 3x^2 + 16y^2 - 48 = 0? Write your answer in simplified, rationalized form. (______,______)

Divide by 48: Now, divide the whole equation by 48 to get the standard form of the ellipse equation. (3x^2)/48 + (16y^2)/48 = 48/48Simplify fractions: Simplify the fractions to get the standard form. x^2/16 + y^2/3 = 1Identify center: Identify the center of the ellipse by looking at the standard form. The standard form is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) is the center. Since there are no (x-h) or (y-k) terms, h and k are both 0. Center of the ellipse: (0, 0)

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What is the center of the ellipse $3x^2 + 16y^2 - 48 = 0$ ? $\newline$ Write your answer in simplified, rationalized form. $\newline$ (,)

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Algebra 2

Find properties of ellipses from equations in general form

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Q. What is the center of the ellipse

3x^2 + 16y^2 - 48 = 0

\newline

Write your answer in simplified, rationalized form.

\newline

(______,______)

Divide by $48$ : Now, divide the whole equation by $48$ to get the standard form of the ellipse equation. $\newline$ $(\frac{3x^2}{48}) + (\frac{16y^2}{48}) = \frac{48}{48}$
Simplify fractions: Simplify the fractions to get the standard form. $\frac{x^2}{16} + \frac{y^2}{3} = 1$
Identify center: Identify the center of the ellipse by looking at the standard form. $\newline$ The standard form is $(x-h)^2/a^2 + (y-k)^2/b^2 = 1$ , where $(h, k)$ is the center. $\newline$ Since there are no $(x-h)$ or $(y-k)$ terms, $h$ and $k$ are both $0$ . $\newline$ Center of the ellipse: $(0, 0)$