What is the center ((y-5)^2}{81}-\frac{(x+8)^2)/9 =1 ?

The center ((y-5)^2}{81}-\frac{(x+8)^2)/9 =1 is (-8,5)

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center ((y-5)^2}{81}-\frac{(x+8)^2)/9 =1

center

\frac{( y - 5 ) ^{2}}{81} - \frac{( x + 8 ) ^{2}}{9} = 1

(- 8, 5)

Solution steps

\frac{( y - 5 ) ^{2}}{81} - \frac{( x + 8 ) ^{2}}{9} = 1

Rewrite

\frac{( y - 5 ) ^{2}}{81} - \frac{( x + 8 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( y - 5 ) ^{2}}{9 ^{2}} - \frac{( x - ( - 8 ) ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 8, 5), a = 9, b = 3

And the center is:

(- 8, 5)

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

3 (x + 2)^{2} - 6 (y - 6)^{2} - 18 = 0

x^{2} - 6 x - 6 = y^{2} + 6 y + 6

y^{2} - 4 y - 4 x^{2} + 13 = 0

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

What is the center ((y-5)^2}{81}-\frac{(x+8)^2)/9 =1 ?
The center ((y-5)^2}{81}-\frac{(x+8)^2)/9 =1 is (-8,5)