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

The center 2x^2+2y^2+4y=0 is (0,-1)

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

center

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

(0, - 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, - 1), a = 1, b = 1

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 1), b = 1, a = 1

And the center is:

(0, - 1)

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

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

\frac{x ^{2}}{9} + y^{2} = 1

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{36} = 1