What is the center (x^2)/4-((y+3)^2)/9 =1 ?

The center (x^2)/4-((y+3)^2)/9 =1 is (0,-3)

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center (x^2)/4-((y+3)^2)/9 =1

center

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

(0, - 3)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} - \frac{( y - ( - 3 ) ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, - 3), a = 2, b = 3

And the center is:

(0, - 3)

16 x^{2} - 9 y^{2} + 64 x + 54 y - 161 = 0

x^{2} - y^{2} < 1

9 x^{2} - 4 y^{2} - 54 x - 8 y + 113 = 0

9 x^{2} - 4 y^{2} = 36

9 x^{2} - 18 x - 24 y - 63 = 4 y^{2}

What is the center (x^2)/4-((y+3)^2)/9 =1 ?
The center (x^2)/4-((y+3)^2)/9 =1 is (0,-3)