What is the center 9x^2+y^2+18x+4y+4=0 ?

The center 9x^2+y^2+18x+4y+4=0 is (-1,-2)

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center 9x^2+y^2+18x+4y+4=0

center

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

(- 1, - 2)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, - 2), a = 1, b = 3

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 1, - 2), b = 3, a = 1

And the center is:

(- 1, - 2)

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

4 x^{2} + 7 y^{2} < 28

\frac{( x + 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{16} = 1

16 x^{2} + 25 y^{2} - 64 x + 150 y - 11 = 0

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

What is the center 9x^2+y^2+18x+4y+4=0 ?
The center 9x^2+y^2+18x+4y+4=0 is (-1,-2)