(16)x^2+(256)x-(361)y^2-3610y+(4273)=0

解答

(16) x^{2} + (256) x - (361) y^{2} - 3610 y + (4273) = 0

(h, k) = (- 8, - 5), a = 34, b = \frac{19 17}{2 2}

求解步骤

16 x^{2} + 256 x - 361 y^{2} - 3610 y + 4273 = 0

将

16 x^{2} + 256 x - 361 y^{2} - 3610 y + 4273 = 0

改写为标准双曲线方程形式

\frac{( y - ( - 5 ) ) ^{2}}{( 34 ) ^{2}} - \frac{( x - ( - 8 ) ) ^{2}}{( \frac{19 17}{2 2} ) ^{2}} = 1

因此双曲线的性质为:

(h, k) = (- 8, - 5), a = 34, b = \frac{19 17}{2 2}

so l v e f or x, x^{2} + 10 x + 25 = x + k (x)^{2}

x^{2} - y^{2} = 280

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

\frac{y - 3}{x - 3} - \frac{2 x + 1}{y + 1} = 0

x^{2} - y^{2} - 8 x - 14 y - 49 = 0