What is the center 12x^2+8x+8y^2+12y=0 ?

The center 12x^2+8x+8y^2+12y=0 is (-1/3 ,-3/4)

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center 12x^2+8x+8y^2+12y=0

center

12 x^{2} + 8 x + 8 y^{2} + 12 y = 0

(- \frac{1}{3}, - \frac{3}{4})

Solution steps

12 x^{2} + 8 x + 8 y^{2} + 12 y = 0

Rewrite

12 x^{2} + 8 x + 8 y^{2} + 12 y = 0

in the form of the standard ellipse equation

\frac{( x - ( - \frac{1}{3} ) ) ^{2}}{( \frac{35}{6 2} ) ^{2}} + \frac{( y - ( - \frac{3}{4} ) ) ^{2}}{( \frac{35}{4 3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{1}{3}, - \frac{3}{4}), a = \frac{35}{6 2}, b = \frac{35}{4 3}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{3}, - \frac{3}{4}), b = \frac{35}{4 3}, a = \frac{35}{6 2}

And the center is:

(- \frac{1}{3}, - \frac{3}{4})

\frac{( x + 5 ) ^{2}}{9} + (y - 1)^{2} = 1

25 x^{2} + 16 y^{2} + 100 x - 96 y = 156

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

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

9 x^{2} + y^{2} = 9

What is the center 12x^2+8x+8y^2+12y=0 ?
The center 12x^2+8x+8y^2+12y=0 is (-1/3 ,-3/4)