3=x^2+4y^2

Solution

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

(h, k) = (0, 0), a = 3, b = \frac{3}{2}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 3, b = \frac{3}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 3, b = \frac{3}{2}

\frac{X ^{2}}{4} + \frac{Y ^{2}}{9} = 4

\frac{x ^{2}}{2.4 8 ^{2}} + \frac{y ^{2}}{2.7 5 ^{2}} = 1

x^{2} + 4 y^{2} - 16 y - 1 = - 1

3 x^{2} + 5 y^{2} - 1 = 0

x^{(2)} + 4 y^{(2)} - 10 x - 40 y + 109 = 0

What is 3=x^2+4y^2 ?
The solution to 3=x^2+4y^2 is Ellipse with (h,k)=(0,0),a=sqrt(3),b=(sqrt(3))/2