What is the center 3x^2+2y^2-12x+8y+19=0 ?

The center 3x^2+2y^2-12x+8y+19=0 is (2,-2)

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center 3x^2+2y^2-12x+8y+19=0

center

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

(2, - 2)

Solution steps

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

Rewrite

3 x^{2} + 2 y^{2} - 12 x + 8 y + 19 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

And the center is:

(2, - 2)

25 x^{2} + y^{2} - 100 x + 8 y + 91 = 0

49 x^{2} + 625 y^{2} = 30625

4 x^{2} + 9 y^{2} - 24 x + 36 y + 36 = 0

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

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

What is the center 3x^2+2y^2-12x+8y+19=0 ?
The center 3x^2+2y^2-12x+8y+19=0 is (2,-2)