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What is the center of the ellipse 3x2+14y2−42=03x^2 + 14y^2 - 42 = 0?\newlineWrite your answer in simplified, rationalized form.\newline((______,,______))

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Q. What is the center of the ellipse 3x2+14y2−42=03x^2 + 14y^2 - 42 = 0?\newlineWrite your answer in simplified, rationalized form.\newline((______,,______))
  1. Divide by 4242: Now, divide the entire equation by 4242 to get the standard form of the ellipse. x2(42/3)+y2(42/14)=1\frac{x^2}{(42/3)} + \frac{y^2}{(42/14)} = 1
  2. Simplify denominators: Simplify the denominators. x214+y23=1\frac{x^2}{14} + \frac{y^2}{3} = 1
  3. Identify center: Identify the center of the ellipse from the standard form.\newlineThe center is at (h,k)(h, k) where the equation is in the form (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2 + (y-k)^2/b^2 = 1.\newlineSince there are no (h,k)(h, k) terms, the center is at (0,0)(0, 0).

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