4=x^2+4y^2

Solution

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{2 ^{2}} + \frac{( y - 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{( y - 1 ) ^{2}}{2} + \frac{5 ( x + 2 ) ^{2}}{3} = 2

2 x^{2} + (y - 1)^{2} \leq 3

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

3 x^{2} + y^{2} - 22 x + 22 y = 6

\frac{x ^{2}}{64} + \frac{y ^{2}}{28} = 1

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