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

Therefore Hyperbola properties are:

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

And the center is:

(6, - 1)

\frac{( y - 5 ) ^{2}}{81} - \frac{( x + 8 ) ^{2}}{9} = 1

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

3 (x + 2)^{2} - 6 (y - 6)^{2} - 18 = 0

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

y^{2} - 4 y - 4 x^{2} + 13 = 0

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)