What is the center 9x^2-4y^2+36x+8y+33=0 ?

The center 9x^2-4y^2+36x+8y+33=0 is (-2,1)

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center 9x^2-4y^2+36x+8y+33=0

center

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

(- 2, 1)

Solution steps

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

Rewrite

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 1), a = \frac{1}{2}, b = \frac{1}{3}

And the center is:

(- 2, 1)

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

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

x^{2} - y^{2} - 2 x - 2 y - 1 = 0

\frac{x ^{2}}{36} - \frac{y ^{2}}{4} = 1

4 x^{2} - 25 y^{2} + 100 y = 200

What is the center 9x^2-4y^2+36x+8y+33=0 ?
The center 9x^2-4y^2+36x+8y+33=0 is (-2,1)