3y^2-x^2+18y+4x+28=0

解答

3 y^{2} - x^{2} + 18 y + 4 x + 28 = 0

(h, k) = (2, - 3), a = 5, b = \frac{5}{3}

求解步骤

3 y^{2} - x^{2} + 18 y + 4 x + 28 = 0

将

3 y^{2} - x^{2} + 18 y + 4 x + 28 = 0

改写为标准双曲线方程形式

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

因此双曲线的性质为:

(h, k) = (2, - 3), a = 5, b = \frac{5}{3}

\frac{50}{x + y} + \frac{8}{x - y} = 7

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

so l v e f or x, \frac{x ^{2}}{4} + 1 = 3 m (x)^{2} - 1

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

(\frac{( x + 1 )}{16})^{2} - (\frac{( y - 4 )}{25})^{2} = 1