center x^2+y^2<= y

Lösung

center

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

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

Schritte zur Lösung

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})

ce n t er 4 x^{2} + 25 y^{2} - 32 x + 200 y + 364 = 0

ce n t er (x 3)^{2} + (y + 4)^{2} = 25

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ce n t er x^{2} + 10 x + 9 y^{2} + 16 = 0

ce n t er 9 x^{2} + 16 y^{2} - 18 x - 64 y - 71 = 0