What is the center 4y^2-x^2+24y+4x+28=0 ?

The center 4y^2-x^2+24y+4x+28=0 is (2,-3)

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center 4y^2-x^2+24y+4x+28=0

center

4 y^{2} - x^{2} + 24 y + 4 x + 28 = 0

(2, - 3)

Solution steps

4 y^{2} - x^{2} + 24 y + 4 x + 28 = 0

Rewrite

4 y^{2} - x^{2} + 24 y + 4 x + 28 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 3), a = 1, b = 2

And the center is:

(2, - 3)

\frac{( y + 6 ) ^{2}}{36} - \frac{( x + 4 ) ^{2}}{48} = 1

y^{2} - 4 x^{2} - 16 x - 2 y - 19 = 0

5 x^{2} - 4 y^{2} - 20 x - 8 y - 4 = 0

0 = 9 y^{2} - x^{2} + 2 x + 72 y + 125

9 x^{2} - 9 y^{2} - 18 x + 54 = 0

What is the center 4y^2-x^2+24y+4x+28=0 ?
The center 4y^2-x^2+24y+4x+28=0 is (2,-3)