What is the center x^2+((y+2)^2)/4 =1 ?

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

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center x^2+((y+2)^2)/4 =1

center

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

(0, - 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, - 2), a = 1, 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 = 1

And the center is:

(0, - 2)

x^{2} + 9 y^{2} - 14 x + 36 y + 49 = 0

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

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

4 x^{2} + 9 y^{2} - 8 x - 36 y + 4 = 0