Question

直线y=ax+1与双曲线3x^2-y^2=1相交于A,B两点,求AB的长

Answer 1

楼上的答案是错误的,AB=2√((1+a^2)(6-a^2))/(3-a^2)中的因子“ 3-a^2”不一定大于0，而且做法极为繁琐，现提供一般方法如下：
y=ax+1代入3x^2-y^2=1
整理得：(3-a^2)x^2-2ax-2=0
令A,B两点的横坐标分别为xA，xB，则xA，xB为上述方程的两根
则由韦达定理：xA+xB=2a/(3-a^2)
xA*xB=-2/(3-a^2)
则|xA-xB|=根号[(xA+xB)^2-4xA*xB]=根号{[2a/(3-a^2)]^2+8/(3-a^2)}
=2/|3-a^2|*根号(6-a^2)
则|AB|=|xA-xB|*根号(1+K^2)=|xA-xB|*根号(1+a^2)=2/|3-a^2|*根号[(6-a^2)(1+a^2)]
(K为直线AB的斜率)
这里的求线段公式|AB|=|xA-xB|*根号(1+K^2)是通用的

Answer 2

把y=ax+1代入3x^2-y^2=1
3x^2-(ax+1)^2=1
(3-a^2)x^2-2ax-2=0

△=(-2a)^2-4(3-a^2)(-2)=4(6-a^2)

x=(2a±√△)/(2(3-a^2))

解得x1=(a+√(6-a^2))/(3-a^2), x2=(a-√(6-a^2))/(3-a^2)

把x1、x2代入y=ax+1

解得y1=(3+a√(6-a^2))/(3-a^2),y2=(3-a√(6-a^2))/(3-a^2)

AB^2=(x1-x2)^2+(y1-y2)^2

=((a+√(6-a^2))/(3-a^2)-(a-√(6-a^2))/(3-a^2))^2
+((3+a√(6-a^2))/(3-a^2)-(3-a√(6-a^2))/(3-a^2))^2

=4*(6-a^2)/(3-a^2)^2+4a^2*(6-a^2)/(3-a^2)^2

=4(-a^4+5a^2+6)/(3-a^2)^2

所以AB的距离为
AB=2√(-a^4+5a^2+6)/(3-a^2)
=2√((1+a^2)(6-a^2))/(3-a^2)

Answer 3

联立消y得AB=2a*根号1+a^2/3-a^2

Answer 4

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