What is the center ((x+4)^2)/9-((y+3)^2)/(16)=1 ?

The center ((x+4)^2)/9-((y+3)^2)/(16)=1 is (-4,-3)

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center ((x+4)^2)/9-((y+3)^2)/(16)=1

center

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

(- 4, - 3)

Solution steps

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 4, - 3)

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1

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

What is the center ((x+4)^2)/9-((y+3)^2)/(16)=1 ?
The center ((x+4)^2)/9-((y+3)^2)/(16)=1 is (-4,-3)