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

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Q. What is the center of the ellipse x2+2y2−12=0x^2 + 2y^2 - 12 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Isolate x and y terms: Now, we need to get the equation in standard form for an ellipse.\newlineDivide both sides by 1212 to isolate the terms with xx and yy.\newlinex212+2y212=1212\frac{x^2}{12} + \frac{2y^2}{12} = \frac{12}{12}
  2. Simplify the equation: Simplify the equation by reducing the fractions. x212+y26=1\frac{x^2}{12} + \frac{y^2}{6} = 1
  3. Identify the center: 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.\newlineOur equation is already in this form with h=0h = 0 and k=0k = 0.\newlineSo, the center of the ellipse is (0,0)(0, 0).

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