What is the center x^2+y^2<= y ?

The center x^2+y^2<= y is (0, 1/2)

Popular >

center x^2+y^2<= y

center

x^{2} + y^{2} \leq y

(0, \frac{1}{2})

Solution steps

x^{2} + y^{2} \leq y

Rewrite

x^{2} + y^{2} \leq y

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

And the center is:

(0, \frac{1}{2})

25 (x - 5)^{2} + 16 (y + 4)^{2} = 400

4 x^{2} + 25 y^{2} - 32 x + 200 y + 364 = 0

(x 3)^{2} + (y + 4)^{2} = 25

\frac{( x - 3 ) ^{2}}{5} + \frac{( y + 2 ) ^{2}}{3} = 1

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0