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

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

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x^2+y^2+2y<= 0

x^{2} + y^{2} + 2 y \leq 0

(h, k) = (0, - 1), a = 1, b = 1

Solution steps

x^{2} + y^{2} + 2 y \leq 0

Rewrite

x^{2} + y^{2} + 2 y \leq 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, - 1), a = 1, b = 1

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 1), b = 1, a = 1

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

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

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{12} = 1

2 x^{2} + y^{2} + 10 x + 12 y = - 12

\frac{x ^{2}}{15} + \frac{y ^{2}}{4} = 1

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