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What is the center of the ellipse 2x2+7y2−28=02x^2 + 7y^2 - 28 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 2x2+7y2−28=02x^2 + 7y^2 - 28 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 2828: Now, divide the entire equation by 2828 to get the standard form of the ellipse equation. (2x2)/28+(7y2)/28=28/28(2x^2)/28 + (7y^2)/28 = 28/28
  2. Simplify fractions: Simplify the fractions. x214+y24=1\frac{x^2}{14} + \frac{y^2}{4} = 1
  3. Identify center: Identify the center of the ellipse by comparing it to the standard form (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.\newlineThe center is at (h,k)=(0,0)(h, k) = (0, 0)

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