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)

(\frac{y - 1}{9})^{2} - (\frac{x - ( - 2 )}{4})^{2} = 1

\frac{( x - 3 ) ^{2}}{a ^{2}} - \frac{( y - 4 ) ^{2}}{b ^{2}} = 1

\frac{( x - 1 ) ^{2}}{9} - \frac{y ^{2}}{4} = 1

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1

\frac{y ^{2}}{25} - x^{2} = 1

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)