What is the center ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 ?

The center ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 is (-3,-5)

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center ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1

center

\frac{( y + 5 ) ^{2}}{16} - \frac{( x + 3 ) ^{2}}{9} = 1

(- 3, - 5)

Solution steps

\frac{( y + 5 ) ^{2}}{16} - \frac{( x + 3 ) ^{2}}{9} = 1

Rewrite

\frac{( y + 5 ) ^{2}}{16} - \frac{( x + 3 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( y - ( - 5 ) ) ^{2}}{4 ^{2}} - \frac{( x - ( - 3 ) ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 3, - 5), a = 4, b = 3

And the center is:

(- 3, - 5)

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

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

- 9 x^{2} + 16 y^{2} - 90 x + 32 y - 353 = 0

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

x^{2} - 2 y^{2} + 3 x + 4 y = 2

What is the center ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 ?
The center ((y+5)^2}{16}-\frac{(x+3)^2)/9 =1 is (-3,-5)