x^2+4y^2=3

Solution

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

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

Solution steps

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

Rewrite

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

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 - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{4} = 1

9 x^{2} + 4 y^{2} - 36 x - 8 y + 4 = 0

x^{2} + 3 y^{2} + 18 y + 18 = 0

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{4} = 1

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

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