What is the center-6x^2+4y^2+6x+16y-1=0 ?

The center-6x^2+4y^2+6x+16y-1=0 is (1/2 ,-2)

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center-6x^2+4y^2+6x+16y-1=0

center

- 6 x^{2} + 4 y^{2} + 6 x + 16 y - 1 = 0

(\frac{1}{2}, - 2)

Solution steps

- 6 x^{2} + 4 y^{2} + 6 x + 16 y - 1 = 0

Rewrite

- 6 x^{2} + 4 y^{2} + 6 x + 16 y - 1 = 0

in the form of a standard hyperbola equation

\frac{( y - ( - 2 ) ) ^{2}}{( \frac{31}{2 2} ) ^{2}} - \frac{( x - \frac{1}{2} ) ^{2}}{( \frac{31}{2 3} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (\frac{1}{2}, - 2), a = \frac{31}{2 2}, b = \frac{31}{2 3}

And the center is:

(\frac{1}{2}, - 2)

y^{2} - 4 x^{2} < 1

- 9 x^{2} + 8 y^{2} + 36 x + 48 y = 36

\frac{y ^{2}}{36} - \frac{x ^{2}}{16} = 1

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

\frac{( y - 1 ) ^{2}}{9} - \frac{( x - 4 ) ^{2}}{25} = 1

What is the center-6x^2+4y^2+6x+16y-1=0 ?
The center-6x^2+4y^2+6x+16y-1=0 is (1/2 ,-2)