What is the center ((x+2)^2)/(36)-((y-3)^2)/(16)=1 ?

The center ((x+2)^2)/(36)-((y-3)^2)/(16)=1 is (-2,3)

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center ((x+2)^2)/(36)-((y-3)^2)/(16)=1

center

\frac{( x + 2 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{16} = 1

(- 2, 3)

Solution steps

\frac{( x + 2 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{16} = 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 3), a = 6, b = 4

And the center is:

(- 2, 3)

x^{2} - y^{2} - 6 x = 7

4 y^{2} - x^{2} = 900

4 x^{2} - 9 y^{2} - 16 x + 18 y - 11 = 0

\frac{( y - 3 ) ^{2}}{11} - \frac{( x + 1 ) ^{2}}{5} = 1

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

What is the center ((x+2)^2)/(36)-((y-3)^2)/(16)=1 ?
The center ((x+2)^2)/(36)-((y-3)^2)/(16)=1 is (-2,3)