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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Rewrite

\frac{x ^{2} + y ^{2} - x}{2 x - x ^{2} - y ^{2}} \geq 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}

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