What is (2y)/(x^2+y^2)<1 ?

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

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(2y)/(x^2+y^2)<1

\frac{2 y}{x ^{2} + y ^{2}} < 1

(h, k) = (0, 1), a = 1, b = 1

Solution steps

\frac{2 y}{x ^{2} + y ^{2}} < 1

Rewrite

\frac{2 y}{x ^{2} + y ^{2}} < 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 1), a = 1, b = 1

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 1), b = 1, a = 1

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{2} = 1

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

9 x^{2} + 25 y^{2} + 18 x - 50 y - 191 = 0

2 x^{2} + 4 y^{2} + 9 x + 9 y - 13 = 0

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

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