y^2<= x^2+y

解答

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

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

求解步骤

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

将

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

改写为标准双曲线方程形式

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

因此双曲线的性质为:

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