3x^2-2y^2+2x-3y+5=0

解答

3 x^{2} - 2 y^{2} + 2 x - 3 y + 5 = 0

(h, k) = (- \frac{1}{3}, - \frac{3}{4}), a = \frac{139}{4 3}, b = \frac{139}{6 2}

求解步骤

3 x^{2} - 2 y^{2} + 2 x - 3 y + 5 = 0

将

3 x^{2} - 2 y^{2} + 2 x - 3 y + 5 = 0

改写为标准双曲线方程形式

\frac{( y - ( - \frac{3}{4} ) ) ^{2}}{( \frac{139}{4 3} ) ^{2}} - \frac{( x - ( - \frac{1}{3} ) ) ^{2}}{( \frac{139}{6 2} ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (- \frac{1}{3}, - \frac{3}{4}), a = \frac{139}{4 3}, b = \frac{139}{6 2}

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

4 x^{2} - 9 y^{2} - 16 x - 36 y - 56 = 0

(x - 5)^{2} - 3 (y + 2)^{2} = 4

\frac{y ^{2}}{8} - \frac{x ^{2}}{5} = 1

7 x^{2} - 24 y^{2} = 41