What is 2y^2+4z^2=4 ?

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

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2y^2+4z^2=4

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

(h, k) = (0, 0), a = 2, b = 1

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 2, b = 1

4 x^{2} + 2 y^{2} - 4 x + 2 y + 1 = 0

\frac{x ^{2}}{2 2 ^{2}} + \frac{y ^{2}}{5 ^{2}} = 1

(x + 5)^{2} + \frac{3}{10} y^{2} = 34

4 - 2 x^{2} - 3 y^{2} = 0

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

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