Question

求以椭圆X^2/16+Y^2/4=1内一点M(1,1)为中点的弦所在的直线方程

Answer 1

设该弦端点为A(x1,y1),B(x2,y2),则
(x1)^2/16+(y1)^2/4=1①
(x2)^2/16+(y2)^2/4=1②
①-②得
(x1-x2)(x1+x2)/16-(y1-y2)(y1+y2)/4=0
因为M(1,1)是AB中点,所以
x1+x2=2,y1+y2=2
所以,(x1-x2)/8-(y1-y2)/2=0
所以,K(L)=K(AB)=(y1-y2)/(x1-x2)=1/4
所以，直线AB为y-1=(1/4)(x-1)
整理得,x-4y+3=0

Answer 2

设经过M的弦与椭圆相交于A、B两点，
A（x1,y1),B(x2,y2),
x1^2/16+y1^2/4=1,(1)
x2^2/16+y2^2/4=1,(2),
(1)-(2)式，
(x1^2-x2^2)/16+(y1^2-y2^2)/4=0,
1/4+(y1-y2)/(x1-x2)*[(y1+y2)/2]/[(x1+x2)/2]=0,(3)
直线方程斜率k=(y1-y2)/(x1-x2),
(y1+y2)/2=1,(x1+x2)=1,
1/4+k*1/1=0,
k=-1/4,
(y-1)/(x-1)=-1/4,
则中点弦的直线方程为：x+4y-5=0.