2=x^2+4y^2

Solution

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

(h, k) = (0, 0), a = 2, b = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 2, b = \frac{1}{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 = \frac{1}{2}

\frac{x ^{2}}{1 ^{2}} + \frac{y ^{2}}{0.4375 ^{2}} = 1

22 x^{2} + 66 y^{2} + 60 x + 4 y - 9 = 0

x^{2} + 2 y^{2} - 5 = 1

\frac{x ^{2}}{841} + \frac{y ^{2}}{484} = 1

\frac{x ^{2}}{1.586 9 ^{2}} + \frac{y ^{2}}{0.9 9 ^{2}} = 1

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