What is x^2+4y^2+4y=3 ?

The solution to x^2+4y^2+4y=3 is Ellipse with (h,k)=(0,-1/2),a=2,b=1

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

49 x^{2} + 16 y^{2} + 196 x - 96 y = 444

\frac{x ^{2}}{16} + \frac{( y - 2 ) ^{2}}{25} = 1

\frac{( x - 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{25} = 1

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{81} = 1

What is x^2+4y^2+4y=3 ?
The solution to x^2+4y^2+4y=3 is Ellipse with (h,k)=(0,-1/2),a=2,b=1