What is the center 321+108x-9x^2+8y+4y^2=0 ?

The center 321+108x-9x^2+8y+4y^2=0 is (6,-1)

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center 321+108x-9x^2+8y+4y^2=0

center

321 + 108 x - 9 x^{2} + 8 y + 4 y^{2} = 0

(6, - 1)

Solution steps

321 + 108 x - 9 x^{2} + 8 y + 4 y^{2} = 0

Rewrite

321 + 108 x - 9 x^{2} + 8 y + 4 y^{2} = 0

in the form of a standard hyperbola equation

\frac{( x - 6 ) ^{2}}{( \frac{641}{3} ) ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{( \frac{641}{2} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (6, - 1), a = \frac{641}{3}, b = \frac{641}{2}

And the center is:

(6, - 1)

2 x^{2} - y^{2} - 2 y = - 5

- 6 x^{2} + 4 y^{2} + 6 x + 16 y - 1 = 0

y^{2} - 4 x^{2} < 1

- 9 x^{2} + 8 y^{2} + 36 x + 48 y = 36

\frac{y ^{2}}{36} - \frac{x ^{2}}{16} = 1

What is the center 321+108x-9x^2+8y+4y^2=0 ?
The center 321+108x-9x^2+8y+4y^2=0 is (6,-1)