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What is the center of the hyperbola x2−4y2−36=0x^2 - 4y^2 - 36 = 0?\newline(_,_)(\_,\_)

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Q. What is the center of the hyperbola x2−4y2−36=0x^2 - 4y^2 - 36 = 0?\newline(_,_)(\_,\_)
  1. Move constant to right: x2−4y2−36=0x^2 - 4y^2 - 36 = 0\newlineMove the constant term to the right side of the equation.\newlinex2−4y2=36x^2 - 4y^2 = 36
  2. Convert to standard form: x2−4y2=36x^2 - 4y^2 = 36\newlineConvert the equation into standard form by dividing both sides by 3636.\newlinex236−4y236=3636\frac{x^2}{36} - \frac{4y^2}{36} = \frac{36}{36}\newlineSimplify the equation.\newlinex236−y29=1\frac{x^2}{36} - \frac{y^2}{9} = 1
  3. Simplify the equation: (x2)/36−(y2)/9=1(x^2)/36 - (y^2)/9 = 1\newlineIdentify the center of the hyperbola.\newlineThe standard form of a hyperbola is (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.\newlineHere, the equation is already in standard form with h=0h = 0 and k=0k = 0.\newlineCenter of the hyperbola: (0,0)(0, 0)

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