What is 2x^2+4y^2-8=0 ?

The solution to 2x^2+4y^2-8=0 is Ellipse with (h,k)=(0,0),a=2,b=sqrt(2)

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2x^2+4y^2-8=0

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

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

Solution steps

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

Rewrite

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

9 (x - 1)^{2} + 4 (y + 2)^{2} = 36

3 x^{2} + 5 y + 5 y^{2} = 5

\frac{( x - 6 ) ^{2}}{36} + \frac{( y + 4 ) ^{2}}{16} = 1

49 x^{2} + 9 y^{2} = 16

\frac{( x - 3 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{32} = 1

What is 2x^2+4y^2-8=0 ?
The solution to 2x^2+4y^2-8=0 is Ellipse with (h,k)=(0,0),a=2,b=sqrt(2)