What is the center x^2+3y^2+2x-12y+10=0 ?

The center x^2+3y^2+2x-12y+10=0 is (-1,2)

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center x^2+3y^2+2x-12y+10=0

center

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

(- 1, 2)

Solution steps

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

Rewrite

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 2), a = 3, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 2), a = 3, b = 1

And the center is:

(- 1, 2)

\frac{x ^{2}}{16} + \frac{( y + 1 ) ^{2}}{4} - 1 = 0

3 x^{2} + 3 y^{2} - 12 y = 0

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

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

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

What is the center x^2+3y^2+2x-12y+10=0 ?
The center x^2+3y^2+2x-12y+10=0 is (-1,2)