Question

求以椭圆x²/8+y²/5=1内的点A为（2,1）为中点的弦所在直线方程

Answer 1

设:弦交椭圆A,B

A(x1,y1) B(x2,y2) x1+x2=2*2=4 y1+y2=2*1=2

x1^2/8+y1^2/5=1

x2^2/8+y2^2/5=1

相减

(x1-x2)(x1+x2)/8+(y1-y2)(y1+y2)/5=0

(x1+x2)/8+(y1+y2)(y1-y2)/(x1-x2)/5=0

4/8+2k/5=0

2k=-5/2

k=-5/4

方程过(2,1)

y-1=-5/4(x-2)