y^2<= x^2-1

解答

y^{2} \leq x^{2} - 1

(h, k) = (0, 0), a = 1, b = 1

求解步骤

y^{2} \leq x^{2} - 1

将

y^{2} \leq x^{2} - 1

改写为标准双曲线方程形式

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

因此双曲线的性质为:

(h, k) = (0, 0), a = 1, b = 1

7 x^{2} - 7 y^{2} + 5 x - 10 y + 22 = 0

\frac{( y - 4 ) ^{2}}{49} - \frac{( x + 5 ) ^{2}}{16} = 1

\frac{( y + 3 ) ^{2}}{9} - \frac{4 ( x + 2 ) ^{2}}{25} = 1

y^{2} - z^{2} = 16

y^{2} = x^{2} + 169