What is (x^2+y^2-x)/(2x-x^2-y^2)>0 ?

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

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(x^2+y^2-x)/(2x-x^2-y^2)>0

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

(h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, 0), b = \frac{1}{2}, a = \frac{1}{2}

(x - 2)^{2} + 9 (y - 1)^{2} = 9

16 x^{2} + 32 x + 9 y^{2} + 72 y + 16 = 0

x^{2} + y^{2} = 12 y

2 x^{2} + y^{2} = 10

4 x^{(2)} + 9 y^{(2)} = 2

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