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

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

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

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

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

Solution steps

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

Rewrite

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

7 x^{2} + 16 y^{2} - 28 x - 128 y + 172 = 0

x^{2} + 5 y^{2} + 6 x - 40 y - 167 = 0

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{4} = 1

4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0

\frac{( x - 2 ) ^{2}}{4} + \frac{( y - 1 ) ^{2}}{9} = 1

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