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What is the center of the ellipse 31x2+2y2−62=031x^2 + 2y^2 - 62 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 31x2+2y2−62=031x^2 + 2y^2 - 62 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide and simplify equation: Now, divide the whole equation by 6262 to get the standard form of the ellipse.x262/31+y262/2=1\frac{x^2}{62/31} + \frac{y^2}{62/2} = 1
  2. Identify standard form: Simplify the denominators. x22+y231=1\frac{x^2}{2} + \frac{y^2}{31} = 1
  3. Determine center 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 x2/2+y2/31=1x^2/2 + y^2/31 = 1, which is the same as (x−0)2/2+(y−0)2/31=1(x-0)^2/2 + (y-0)^2/31 = 1.\newlineSo, the center is (0,0)(0, 0).

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