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What is the center of the ellipse 3x2+2y2−84=03x^2 + 2y^2 - 84 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 3x2+2y2−84=03x^2 + 2y^2 - 84 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide and Simplify: Now, divide the entire equation by 8484 to get the standard form of the ellipse.\newline3x284+2y284=8484 \frac{3x^2}{84} + \frac{2y^2}{84} = \frac{84}{84} \newlineSimplify the fractions.\newlinex2/28+y2/42=1 x^2/28 + y^2/42 = 1
  2. Standard Form of Ellipse: The standard form of an ellipse is (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k)(h, k) is the center.\newlineHere, we have x2/28+y2/42=1x^2/28 + y^2/42 = 1, which can be written as (x−0)2/28+(y−0)2/42=1(x-0)^2/28 + (y-0)^2/42 = 1.\newlineSo, the center of the ellipse is (0,0)(0, 0).

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