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

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

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

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

(h, k) = (- 2, 0), a = 22, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 0), a = 22, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 0), a = 22, b = 2

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

2 = 4 x^{2} + y^{2} + 1

9 x^{2} + 16 y^{2} + 54 x - 32 y - 47 = 0

16 x^{2} + 25 y^{2} - 32 x - 150 y - 159 = 0

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

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