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

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

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

center

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

(- 3, 4)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 3, 4)

18 x^{2} - 32 y^{2} = 2

16 x^{2} - 96 x - y^{2} - 2 y + 139 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{64} = 1

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

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

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