What is the center (x^2)/4-((y+1)^2)/(16)=1 ?

The center (x^2)/4-((y+1)^2)/(16)=1 is (0,-1)

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center (x^2)/4-((y+1)^2)/(16)=1

center

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

(0, - 1)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{4 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, - 1), a = 2, b = 4

And the center is:

(0, - 1)

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

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1

x^{2} - 4 y^{2} - 20 x - 64 y - 172 = 0

- 9 x^{2} - 32 y = - 16 y^{2} + 128

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

What is the center (x^2)/4-((y+1)^2)/(16)=1 ?
The center (x^2)/4-((y+1)^2)/(16)=1 is (0,-1)