y^2+4z^2=4

Solution

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

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

Solution steps

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

Rewrite

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

x^{2} + 4 y^{2} - 2 x + 16 y + 13 = 0

36 x^{2} + 9 y^{2} = 1

4 x^{2} + 9 y^{2} - 40 x + 54 y + 145 = 0

\frac{( y + 4 ) ^{2}}{9} + \frac{( x - 9 ) ^{2}}{25} = 1

\frac{( x - 100 ) ^{2}}{400} + \frac{( y - 90 ) ^{2}}{900} = 1

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