What is the center-4x^2+y^2+16x+12y+16=0 ?

The center-4x^2+y^2+16x+12y+16=0 is (2,-6)

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center-4x^2+y^2+16x+12y+16=0

center

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

(2, - 6)

Solution steps

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

Rewrite

- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (2, - 6), a = 2, b = 1

And the center is:

(2, - 6)

36 x^{2} - y^{2} - 216 x + 10 y + 263 = 0

4 x^{2} - 9 y^{2} - 24 x - 36 y - 36 = 0

x^{2} - y^{2} = 6

x^{2} - 25 y^{2} - 14 x + 100 y - 76 = 0

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1

What is the center-4x^2+y^2+16x+12y+16=0 ?
The center-4x^2+y^2+16x+12y+16=0 is (2,-6)