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What is the center of the ellipse 3x2+14y2−84=03x^2 + 14y^2 - 84 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 3x2+14y2−84=03x^2 + 14y^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. \newlinex284/3+y284/14=1\frac{x^2}{84/3} + \frac{y^2}{84/14} = 1
  2. Simplify denominators: Simplify the denominators. x228+y26=1\frac{x^2}{28} + \frac{y^2}{6} = 1
  3. Identify center: Identify the center of the ellipse from the standard form.\newlineThe center is at (h,k)(h, k) where hh and kk are the coordinates that the xx and yy are subtracted from in the equation. Since there is no subtraction, hh and kk are both 00.\newlineCenter of the ellipse: (0,0)(0, 0)

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