What is the center x^2+y^2-6y=0 ?

The center x^2+y^2-6y=0 is (0,3)

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center x^2+y^2-6y=0

center

x^{2} + y^{2} - 6 y = 0

(0, 3)

Solution steps

x^{2} + y^{2} - 6 y = 0

Rewrite

x^{2} + y^{2} - 6 y = 0

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{3 ^{2}} + \frac{( y - 3 ) ^{2}}{3 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 3), a = 3, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 3), b = 3, a = 3

And the center is:

(0, 3)

4 (x + 5)^{2} + 8 (y - 2)^{2} = 32

4 x^{2} + 9 y^{2} + 24 x + 18 y + 9 = 0

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

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

\frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1