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

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

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center 2x^2+3y^2+4x=1

center

2 x^{2} + 3 y^{2} + 4 x = 1

(- 1, 0)

Solution steps

2 x^{2} + 3 y^{2} + 4 x = 1

Rewrite

2 x^{2} + 3 y^{2} + 4 x = 1

in the form of the standard ellipse equation

\frac{( x - ( - 1 ) ) ^{2}}{( \frac{3}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), a = \frac{3}{2}, b = 1

And the center is:

(- 1, 0)

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

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

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

x^{2} + 3 y^{2} = 1

25 x^{2} + 16 y^{2} = 400