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What is the center of the ellipse 3x2+32y2−96=03x^2 + 32y^2 - 96 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 3x2+32y2−96=03x^2 + 32y^2 - 96 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 9696: Divide the entire equation by 9696 to get the standard form of the ellipse equation.\newline(3x2)/96+(32y2)/96=96/96(3x^2)/96 + (32y^2)/96 = 96/96\newlineSimplify the fractions.\newlinex2/32+y2/3=1x^2/32 + y^2/3 = 1
  2. Simplify fractions: Identify the center of the ellipse from the standard form equation.\newlineThe 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, the equation is already in the form (x−0)2/32+(y−0)2/3=1(x-0)^2/32 + (y-0)^2/3 = 1.\newlineSo, the center is (0,0)(0, 0).

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