What is the center 16y^2-x^2+2x+64y+47=0 ?

The center 16y^2-x^2+2x+64y+47=0 is (1,-2)

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center 16y^2-x^2+2x+64y+47=0

center

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

(1, - 2)

Solution steps

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

Rewrite

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (1, - 2), a = 1, b = 4

And the center is:

(1, - 2)

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0

- 9 x^{2} + 4 y^{2} - 36 x - 16 y - 164 = 0

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

11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0

16 x^{2} - 9 y^{2} - 144 = 0

What is the center 16y^2-x^2+2x+64y+47=0 ?
The center 16y^2-x^2+2x+64y+47=0 is (1,-2)