What is (2y)/(1-x^2-y^2)= 1/2 ?

The solution to (2y)/(1-x^2-y^2)= 1/2 is Ellipse with (h,k)=(0,-2),a=sqrt(5),b=sqrt(5)

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(2y)/(1-x^2-y^2)= 1/2

\frac{2 y}{1 - x ^{2} - y ^{2}} = \frac{1}{2}

(h, k) = (0, - 2), a = 5, b = 5

Solution steps

\frac{2 y}{1 - x ^{2} - y ^{2}} = \frac{1}{2}

Rewrite

\frac{2 y}{1 - x ^{2} - y ^{2}} = \frac{1}{2}

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( 5 ) ^{2}} + \frac{( y - ( - 2 ) ) ^{2}}{( 5 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, - 2), a = 5, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, - 2), b = 5, a = 5

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

4 x^{2} + y^{2} - 16 x + 6 y - 75 = 0

x^{2} + 5 y^{2} - 4 x + 40 y + 64 = 0

4 x^{2} - 24 x + \frac{y ^{2}}{16} + \frac{y}{8} + \frac{573}{16} = 0

\frac{( x + 2 ) ^{2}}{4} + \frac{( y - 7 ) ^{2}}{49} = 1

What is (2y)/(1-x^2-y^2)= 1/2 ?
The solution to (2y)/(1-x^2-y^2)= 1/2 is Ellipse with (h,k)=(0,-2),a=sqrt(5),b=sqrt(5)