Question

1.已知双曲线x^2-y^2=2,求以M(3,1)为中点的弦所在直线方程.

2.求过椭圆x^2/25+y^2/9=1的焦点，倾斜角为45°的弦AB的长.

Answer 1

1,M(3,1)为中点的弦AB:
xA+xB=2xM=2*2=4,yA+yB=2
k(AB)=(yA-yB)/(xA-xB)=(y-1)/(x-3)
[(xA)^2-(yA)]^2-[(xB)^2-(yB)^2]=2-2=0
[(xA)^2-(xB)]^2-[(yA)^2-(yB)^2]=0
(xA+xB)*(xA-xB)-(yA+yB)*(yA-yB)=0
(xA+xB) -(yA+yB)*(yA-yB)/ (xA-xB) =0
4-2(y-1)/(x-3)=0
弦所在直线方程:2x-y-5=0
2,a^2=25,b^2=9,c^2=16,c=4
F1(-4,0),F2(4,0)
k(AB)=1
直线AB有两条:1，y=x-4；2，y=x+4
到这应该会了啦