1=x^2+3y^2

Solution

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

(h, k) = (0, 0), a = 1, b = \frac{1}{3}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 1, b = \frac{1}{3}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 1, b = \frac{1}{3}

4 x^{2} + y^{2} - 24 x + 2 y + 21 = 0

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

9 x^{2} + 4 y^{2} + 18 x - 40 y + 73 = 0

\frac{( x + 8 ) ^{2}}{64} + \frac{( y - 11 ) ^{2}}{36} = 1

x^{2} + 2 x + 4 y^{2} = 0

What is 1=x^2+3y^2 ?
The solution to 1=x^2+3y^2 is Ellipse with (h,k)=(0,0),a=1,b=sqrt(1/3)