x^2+4z^2=4

Solution

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 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

\frac{( x - 2 ) ^{2}}{25} + \frac{( v + 3 ) ^{2}}{9} = 1

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{8 ^{2}} = 1

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

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

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 1 ) ^{2}}{100} = 1

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