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

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)