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What is the center of the ellipse x2+12y2−24=0x^2 + 12y^2 - 24 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse x2+12y2−24=0x^2 + 12y^2 - 24 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 2424: Now, we need to get the equation in standard form for an ellipse.\newlineDivide both sides by 2424 to get the coefficients of x2x^2 and y2y^2 to be 11.\newlinex224+12y224=2424\frac{x^2}{24} + \frac{12y^2}{24} = \frac{24}{24}\newlinex224+y22=1\frac{x^2}{24} + \frac{y^2}{2} = 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.\newlineFrom our equation x2/24+y2/2=1x^2/24 + y^2/2 = 1, we can see that h=0h = 0 and k=0k = 0.\newlineSo, the center of the ellipse is (0,0)(0, 0).

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