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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

\frac{( y - 3 ) ^{2}}{16} + \frac{( x + 2 ) ^{2}}{9} = 1

7 (x - 2004)^{2} = 23 - y^{2}

\frac{( x - 6 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{9} = 1

x^{2} + 2 y^{2} + 8 x - 4 y - 6 = 0

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