What is x^2+y^2-2y>0 ?

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

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x^2+y^2-2y>0

x^{2} + y^{2} - 2 y > 0

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

Solution steps

x^{2} + y^{2} - 2 y > 0

Rewrite

x^{2} + y^{2} - 2 y > 0

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 - 1 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

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

25 x^{(2)} + 16 y^{(2)} - 96 y - 156 = 0

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} = 1

5 x^{2} + 20 y^{2} - 25 x - 10 y = 5

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