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What is the center of the ellipse 3x2+16y2−48=03x^2 + 16y^2 - 48 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)

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Q. What is the center of the ellipse 3x2+16y2−48=03x^2 + 16y^2 - 48 = 0?\newlineWrite your answer in simplified, rationalized form.\newline(______,______)
  1. Divide by 4848: Now, divide the whole equation by 4848 to get the standard form of the ellipse equation.\newline(3x248)+(16y248)=4848(\frac{3x^2}{48}) + (\frac{16y^2}{48}) = \frac{48}{48}
  2. Simplify fractions: Simplify the fractions to get the standard form. x216+y23=1\frac{x^2}{16} + \frac{y^2}{3} = 1
  3. Identify center: Identify the center of the ellipse by looking at the standard form.\newlineThe standard form 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.\newlineSince there are no (x−h)(x-h) or (y−k)(y-k) terms, hh and kk are both 00.\newlineCenter of the ellipse: (0,0)(0, 0)

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