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

The solution to x^2+4y^2+7=12y-2x is Ellipse with (h,k)=(-1, 3/2),a=sqrt(3),b=(sqrt(3))/2

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

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

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

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

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

x^{2} + 2 y^{2} + 2 x - 20 y + 43 = 0

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

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

What is x^2+4y^2+7=12y-2x ?
The solution to x^2+4y^2+7=12y-2x is Ellipse with (h,k)=(-1, 3/2),a=sqrt(3),b=(sqrt(3))/2