What is the center 4x^2-24x-25y^2+250y-489=0 ?

The center 4x^2-24x-25y^2+250y-489=0 is (3,5)

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center 4x^2-24x-25y^2+250y-489=0

center

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

(3, 5)

Solution steps

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

Rewrite

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, 5), a = 2, b = 5

And the center is:

(3, 5)

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

\frac{( y - 5 ) ^{2}}{16} - \frac{( x - 4 ) ^{2}}{64} = 1

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y - 2 = 0

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

9 x^{2} - 18 x - 24 y - 63 = 4 y^{2}

What is the center 4x^2-24x-25y^2+250y-489=0 ?
The center 4x^2-24x-25y^2+250y-489=0 is (3,5)