What is x^2+y^2+4y=0 ?

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

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

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

(h, k) = (0, - 2), a = 2, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, - 2), a = 2, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 2), b = 2, a = 2

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{9} = 1

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

x^{2} + y^{2} - 6 x - 8 y = 0

y, y^{2} - x^{2} + 4 x + 8 y + 12 = 0

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