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What is the center of the ellipse 12x2+7y2−84=012x^2 + 7y^2 - 84 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 12x2+7y2−84=012x^2 + 7y^2 - 84 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 8484: Now, divide the entire equation by 8484 to get the standard form of the ellipse equation.\newline(12x284)+(7y284)=8484(\frac{12x^2}{84}) + (\frac{7y^2}{84}) = \frac{84}{84}
  2. Simplify fractions: Simplify the fractions. x27+y212=1\frac{x^2}{7} + \frac{y^2}{12} = 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. The center (h,k)(h,k) is at (0,0)(0,0) since there are no terms to shift the ellipse left/right or up/down.

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