What is the center 4x^2-25y^2+16x+50y-109=0 ?

The center 4x^2-25y^2+16x+50y-109=0 is (-2,1)

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center 4x^2-25y^2+16x+50y-109=0

center

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

(- 2, 1)

Solution steps

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

Rewrite

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (- 2, 1), a = 5, b = 2

And the center is:

(- 2, 1)

\frac{y ^{2}}{16} - \frac{x ^{2}}{9} = 1

\frac{x ^{2}}{4} - \frac{( y + 1 ) ^{2}}{16} = 1

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

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

x^{2} - 4 y^{2} - 20 x - 64 y - 172 = 0

What is the center 4x^2-25y^2+16x+50y-109=0 ?
The center 4x^2-25y^2+16x+50y-109=0 is (-2,1)