What is the center ((x-5)^2)/7-((y+3)^2)/9 =1 ?

The center ((x-5)^2)/7-((y+3)^2)/9 =1 is (5,-3)

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center ((x-5)^2)/7-((y+3)^2)/9 =1

center

\frac{( x - 5 ) ^{2}}{7} - \frac{( y + 3 ) ^{2}}{9} = 1

(5, - 3)

Solution steps

\frac{( x - 5 ) ^{2}}{7} - \frac{( y + 3 ) ^{2}}{9} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{7} - \frac{( y + 3 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

\frac{( x - 5 ) ^{2}}{( 7 ) ^{2}} - \frac{( y - ( - 3 ) ) ^{2}}{3 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (5, - 3), a = 7, b = 3

And the center is:

(5, - 3)

3 x^{2} - 3 y^{2} - 12 x - 6 y + 3 = 0

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

2 y^{2} - 5 x^{2} - 56 y - 90 x - 807 = 0

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

\frac{x ^{2}}{0.0008} - \frac{y ^{2}}{0.999} = 1

What is the center ((x-5)^2)/7-((y+3)^2)/9 =1 ?
The center ((x-5)^2)/7-((y+3)^2)/9 =1 is (5,-3)