What is the center ((x-3)^2)/(25)-((y+4)^2)/(36)=1 ?

The center ((x-3)^2)/(25)-((y+4)^2)/(36)=1 is (3,-4)

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center ((x-3)^2)/(25)-((y+4)^2)/(36)=1

center

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{36} = 1

(3, - 4)

Solution steps

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{36} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{36} = 1

in the form of a standard hyperbola equation

\frac{( x - 3 ) ^{2}}{5 ^{2}} - \frac{( y - ( - 4 ) ) ^{2}}{6 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (3, - 4), a = 5, b = 6

And the center is:

(3, - 4)

7 x^{2} - 9 y^{2} - 14 x + 72 y - 200 = 0

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

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

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

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{64} = 1

What is the center ((x-3)^2)/(25)-((y+4)^2)/(36)=1 ?
The center ((x-3)^2)/(25)-((y+4)^2)/(36)=1 is (3,-4)