What is the center 3x^2+y^2+2x-6y=3 ?

The center 3x^2+y^2+2x-6y=3 is (-1/3 ,3)

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center 3x^2+y^2+2x-6y=3

center

3 x^{2} + y^{2} + 2 x - 6 y = 3

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

Solution steps

3 x^{2} + y^{2} + 2 x - 6 y = 3

Rewrite

3 x^{2} + y^{2} + 2 x - 6 y = 3

in the form of the standard ellipse equation

\frac{( x - ( - \frac{1}{3} ) ) ^{2}}{( \frac{37}{3} ) ^{2}} + \frac{( y - 3 ) ^{2}}{( \frac{37}{3} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- \frac{1}{3}, 3), a = \frac{37}{3}, b = \frac{37}{3}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- \frac{1}{3}, 3), b = \frac{37}{3}, a = \frac{37}{3}

And the center is:

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

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

x^{2} + 9 y^{2} - 14 x + 36 y + 49 = 0

\frac{( x - 1 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{25} = 1

x^{2} + 4 y^{2} + 8 y = 0

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