4=x^2+2y^2

Solution

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

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

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

\frac{x ^{2}}{24} + \frac{y ^{2}}{20} = 1

16 (x^{2}) + 9 (y^{2}) - 144 = 0

49 x^{2} + 16 y^{2} + 98 x - 64 y - 671 = 0

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