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

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

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

x^{2} + 2 y + y^{2} < 0

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

Solution steps

x^{2} + 2 y + y^{2} < 0

Rewrite

x^{2} + 2 y + y^{2} < 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

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

64 y^{2} + 300 x + 100 x^{2} = - 81 + 256 y

x^{2} + y^{2} - 9 - 4 x^{2} - 9 y^{2} - 36 = 0

- 16 x^{2} - 9 y^{2} = 144

7 x^{2} + 3 y^{2} = 55

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