What is the center x^2+4y^2+7=12y-2x ?

The center x^2+4y^2+7=12y-2x is (-1, 3/2)

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center x^2+4y^2+7=12y-2x

center

x^{2} + 4 y^{2} + 7 = 12 y - 2 x

(- 1, \frac{3}{2})

Solution steps

x^{2} + 4 y^{2} + 7 = 12 y - 2 x

Rewrite

x^{2} + 4 y^{2} + 7 = 12 y - 2 x

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

And the center is:

(- 1, \frac{3}{2})

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

9 (x - 3)^{2} + (y + 2)^{2} = 18

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

2 x^{2} + y^{2} = 50

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

What is the center x^2+4y^2+7=12y-2x ?
The center x^2+4y^2+7=12y-2x is (-1, 3/2)