Question

过椭圆x^2/16+y^2/4=1内一点M（2,1）引一直线与椭圆交于A,B两点若弦AB被M平分,M所在的直线是?

Answer 1

设:直线y=kx+b

将M（2,1)代入,得:

1=2k+b

b=1-2k

所以,直线为y=kx+1-2k,

代入x^2/16+y^2/4=1,得:x^2/16+(kx+1-2k)^2/4=1

x^2+4(kx+1-2k)^2=16

(1+4k^2)x^2+8k(1-2k)x+4(1-2k)^2-16=0

x1+x2=2*2=4

-8k(1-2k)/(1+4k^2)=4

k=-2

所以,直线为y=kx+1-2k=-2x+5