What is the center ((x-2)^2)/4-((y+2)^2)/5 =1 ?

The center ((x-2)^2)/4-((y+2)^2)/5 =1 is (2,-2)

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center ((x-2)^2)/4-((y+2)^2)/5 =1

center

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

(2, - 2)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 2), a = 2, b = 5

And the center is:

(2, - 2)

+ \frac{x ^{2}}{49} - \frac{y ^{2}}{9} = 1

5 x^{2} - 2 y^{2} + 10 x - 12 y = 23

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

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0

What is the center ((x-2)^2)/4-((y+2)^2)/5 =1 ?
The center ((x-2)^2)/4-((y+2)^2)/5 =1 is (2,-2)