What is the center 3x^2+3y^2-12y=0 ?

The center 3x^2+3y^2-12y=0 is (0,2)

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center 3x^2+3y^2-12y=0

center

3 x^{2} + 3 y^{2} - 12 y = 0

(0, 2)

Solution steps

3 x^{2} + 3 y^{2} - 12 y = 0

Rewrite

3 x^{2} + 3 y^{2} - 12 y = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - 2 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 2), a = 2, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 2), b = 2, a = 2

And the center is:

(0, 2)

\frac{( x - 4 ) ^{2}}{3 ^{2}} + \frac{( y - 2 ) ^{2}}{4 ^{2}} = 1

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

\frac{( x - 2 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1

\frac{( x - 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{9} = 1

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0