What is the center 2x^2+3y^2-4x+6y-1=0 ?

The center 2x^2+3y^2-4x+6y-1=0 is (1,-1)

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center 2x^2+3y^2-4x+6y-1=0

center

2 x^{2} + 3 y^{2} - 4 x + 6 y - 1 = 0

(1, - 1)

Solution steps

2 x^{2} + 3 y^{2} - 4 x + 6 y - 1 = 0

Rewrite

2 x^{2} + 3 y^{2} - 4 x + 6 y - 1 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(1, - 1)

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

4 x^{2} + 25 y^{2} = 100

2 (x + \frac{1}{4})^{2} + y^{2} = 8

x^{2} + 4 y^{2} = 16

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

What is the center 2x^2+3y^2-4x+6y-1=0 ?
The center 2x^2+3y^2-4x+6y-1=0 is (1,-1)