What is the center ((x-3)^2)/(49)-((y+5)^2)/(25)=1 ?

The center ((x-3)^2)/(49)-((y+5)^2)/(25)=1 is (3,-5)

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center ((x-3)^2)/(49)-((y+5)^2)/(25)=1

center

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1

(3, - 5)

Solution steps

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(3, - 5)

25 x^{2} - 9 y^{2} = 225

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

18 x^{2} - 32 y^{2} - 108 y - 128 x - 254 = 0

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0

What is the center ((x-3)^2)/(49)-((y+5)^2)/(25)=1 ?
The center ((x-3)^2)/(49)-((y+5)^2)/(25)=1 is (3,-5)