What is the center 25x^2-16y^2+50x-32y-391=0 ?

The center 25x^2-16y^2+50x-32y-391=0 is (-1,-1)

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center 25x^2-16y^2+50x-32y-391=0

center

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

(- 1, - 1)

Solution steps

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

Rewrite

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

in the form of a standard hyperbola equation

\frac{( x - ( - 1 ) ) ^{2}}{4 ^{2}} - \frac{( y - ( - 1 ) ) ^{2}}{5 ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 1, - 1), a = 4, b = 5

And the center is:

(- 1, - 1)

x^{2} - 2 y^{2} + 3 x + 4 y = 2

\frac{( x - 5 ) ^{2}}{4} - \frac{( y - 3 ) ^{2}}{9} = 1

4 y^{2} - 9 x^{2} + 8 y - 54 x - 113 = 0

y^{2} - x^{2} = \frac{9}{25}

9 x^{2} - 16 y^{2} - 108 x + 128 y + 212 = 0

What is the center 25x^2-16y^2+50x-32y-391=0 ?
The center 25x^2-16y^2+50x-32y-391=0 is (-1,-1)