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

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

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

center

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

(- 2, 3)

Solution steps

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

Rewrite

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

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 2, 3)

\frac{( x + 4 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

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

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