设点A(X1,y1)B(X2,y2)在双曲线上
则X1^2-4y1^2=4 ①
X2^2-4y2^2=4 ②
① - ②得:(X1+X2)(X1-X2)-4(y1+y2)(y1-y2)=0
则k=y1-y2/x1-x2=x1+x2/4(y1-y2)
p为AB的中点X1+X2=8
y1+y2=2
解得k=2
y-1=2(X-4)
所以直线l:2x-y-7=0
设点A(X1,y1)B(X2,y2)在双曲线上
则X1^2-4y1^2=4 ①
X2^2-4y2^2=4 ②
① - ②得:(X1+X2)(X1-X2)-4(y1+y2)(y1-y2)=0
则k=y1-y2/x1-x2=x1+x2/4(y1-y2)
p为AB的中点X1+X2=8
y1+y2=2
解得k=2
y-1=2(X-4)
所以直线l:2x-y-7=0