Question

经过点M(2,1)做直线交双曲线X^2-Y^2=1于A,B两点,如果点M为线段AB的中点,求直线AB的方程.

Answer 1

y-1=k(x-2)

y=kx+(1-2k)

代入x²-y²=1

x²-[k²x²+2k(1-2k)x+(1-2k)²]=1

(1-k²)x²-2k(1-2k)x-(1-2k)²-1=0

x1+x2=2k(1-2k)/(1-k²)

中点横坐标是(x1+x2)/2=k(1-2k)/(1-k²)

所以k(1-2k)/(1-k²)=2

k-2k²=2-2k²

k=2

所以是2x-y-3=0