What is the center x^2+y^2+4y=0 ?

The center x^2+y^2+4y=0 is (0,-2)

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center x^2+y^2+4y=0

center

x^{2} + y^{2} + 4 y = 0

(0, - 2)

Solution steps

x^{2} + y^{2} + 4 y = 0

Rewrite

x^{2} + y^{2} + 4 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)

2 x^{2} + y^{2} = 50

4 x^{2} + y^{2} = 16

\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{9} = 1