What is the center-4x^2+24x+16y^2-128y+156=0 ?

The center-4x^2+24x+16y^2-128y+156=0 is (3,4)

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center-4x^2+24x+16y^2-128y+156=0

center

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

(3, 4)

Solution steps

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

Rewrite

- 4 x^{2} + 24 x + 16 y^{2} - 128 y + 156 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, 4), a = 2, b = 4

And the center is:

(3, 4)

\frac{( x - 1 ) ^{2}}{9} - \frac{( y + 6 ) ^{2}}{4} = 1

25 x^{2} - 4 y^{2} - 100 x - 16 y - 16 = 0

9 x^{2} - 16 y^{2} - 36 x + 96 y + 252 = 0

5 y^{2} - 16 x^{2} + 400 = 0

4 x^{2} - 9 y^{2} - 16 x + 18 y - 7 = 0

What is the center-4x^2+24x+16y^2-128y+156=0 ?
The center-4x^2+24x+16y^2-128y+156=0 is (3,4)