What is the center-((x-3)^2}{25}+\frac{(y+2)^2)/9 =1 ?

The center-((x-3)^2}{25}+\frac{(y+2)^2)/9 =1 is (3,-2)

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center-((x-3)^2}{25}+\frac{(y+2)^2)/9 =1

center

- \frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{9} = 1

(3, - 2)

Solution steps

- \frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{9} = 1

Rewrite

- \frac{( x - 3 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{9} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, - 2), a = 3, b = 5

And the center is:

(3, - 2)

9 y^{2} - x^{2} + 2 x + 54 y + 62 = 0

x^{2} - 4 y^{2} + 8 x + 16 y - 4 = 0

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

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 5 ) ^{2}}{9} = 1

x^{2} - y^{2} + 2 y = 0

What is the center-((x-3)^2}{25}+\frac{(y+2)^2)/9 =1 ?
The center-((x-3)^2}{25}+\frac{(y+2)^2)/9 =1 is (3,-2)