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What is the center of the ellipse x2+36y2−72=0x^2 + 36y^2 - 72 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse x2+36y2−72=0x^2 + 36y^2 - 72 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 7272: Divide the entire equation by 7272 to get the standard form of the ellipse equation.\newlinex272+36y272=7272\frac{x^2}{72} + \frac{36y^2}{72} = \frac{72}{72}\newlinex272+y22=1\frac{x^2}{72} + \frac{y^2}{2} = 1
  2. Identify center: Identify the center of the ellipse by comparing the equation to the standard form (x−h)2/a2+(y−k)2/b2=1(x-h)^2/a^2 + (y-k)^2/b^2 = 1. Here, hh and kk are both 00, so the center is at (0,0)(0, 0).

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