Question

过点p(1,1)作椭圆c:16/x^2+4/y^2=1的弦ab,求点a.b所在直线l的方程

Answer 1

按题目条件,有无数条直线即：y-1=k*(x-1)

题目改为P是AB的中点,解法如下

xA+xB=2xP=2,yA+yB=2yP=2

[(xA)^2/16+(yA)^2/4]-[(xB)^2/16+(yB)^2/4]=1-1

(xA+xB)*(xA-xB)+4(yA+yB)*(yA-yB)=0

(xA+xB)+4(yA+yB)*(yA-yB)/(xA-xB)=0

2+4*2*(y-1)/(x-1)=0

x+4y-5=0