What is the center ((y+3)^2)/4-((x+1)^2)/(16)=1 ?

The center ((y+3)^2)/4-((x+1)^2)/(16)=1 is (-1,-3)

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center ((y+3)^2)/4-((x+1)^2)/(16)=1

center

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

(- 1, - 3)

Solution steps

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

Rewrite

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 1, - 3), a = 2, b = 4

And the center is:

(- 1, - 3)

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1

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

4 y^{2} - 9 x^{2} + 16 y + 18 x = 29

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

\frac{x ^{2}}{25} - \frac{y ^{2}}{49} = 1

What is the center ((y+3)^2)/4-((x+1)^2)/(16)=1 ?
The center ((y+3)^2)/4-((x+1)^2)/(16)=1 is (-1,-3)