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What is the center of the ellipse x2+2y2−16=0x^2 + 2y^2 - 16 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(_____,_____)(\_\_\_\_\_,\_\_\_\_\_)

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Q. What is the center of the ellipse x2+2y2−16=0x^2 + 2y^2 - 16 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(_____,_____)(\_\_\_\_\_,\_\_\_\_\_)
  1. Move constant term: Move the constant term to the right side to start getting the equation into standard form.\newlinex2+2y2=16x^2 + 2y^2 = 16
  2. Divide by 1616: Divide the equation by 1616 to get the standard form of the ellipse equation. x216+y28=1\frac{x^2}{16} + \frac{y^2}{8} = 1
  3. Identify center: Identify the center of the ellipse by comparing 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.\newlineCenter is (0,0)(0, 0) since there's no hh and kk in the equation.

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