What is the center 25x^2-16y^2+50x+128y-631=0 ?

The center 25x^2-16y^2+50x+128y-631=0 is (-1,4)

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center 25x^2-16y^2+50x+128y-631=0

center

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0

(- 1, 4)

Solution steps

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0

Rewrite

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

And the center is:

(- 1, 4)

- 9 x^{2} + 4 y^{2} - 36 x - 16 y - 164 = 0

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

11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0

16 x^{2} - 9 y^{2} - 144 = 0

x^{2} - 3 y^{2} - 2 x - 6 y + 7 = 0

What is the center 25x^2-16y^2+50x+128y-631=0 ?
The center 25x^2-16y^2+50x+128y-631=0 is (-1,4)