渐近线 4x^2-y^2+40x-4y+80=0

解答

渐近线

4 x^{2} - y^{2} + 40 x - 4 y + 80 = 0

y = 2 (x + 5) - 2, y = - 2 (x + 5) - 2

求解步骤

y = \pm \frac{b}{a} (x - h) + k

计算双曲线性质

4 x^{2} - y^{2} + 40 x - 4 y + 80 = 0 :

左右开口双曲线，

(h, k) = (- 5, - 2), a = 2, b = 4

y = \frac{4}{2} (x - (- 5)) + (- 2), y = - \frac{4}{2} (x - (- 5)) + (- 2)

整理后得

y = 2 (x + 5) - 2, y = - 2 (x + 5) - 2

a sy m pt o t es x^{2} - 16 y^{2} = 16

a sy m pt o t es - \frac{y ^{2}}{1} + \frac{x ^{2}}{49} = 1

a sy m pt o t es \frac{x ^{2}}{9} - \frac{y ^{2}}{49} = 1

ce n t er \frac{( x - 3 ) ^{2}}{49} - \frac{( y + 5 ) ^{2}}{25} = 1

ce n t er 25 x^{2} - 9 y^{2} = 225