What is the center x^2+4x+4y^2-8y+4=0 ?

The center x^2+4x+4y^2-8y+4=0 is (-2,1)

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center x^2+4x+4y^2-8y+4=0

center

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

(- 2, 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 1), a = 2, b = 1

And the center is:

(- 2, 1)

\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

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

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0