Question

过A(2,1)引直线与椭圆x2/16+y2/9=1相交于P,Q两点,若A恰好是PQ的中点,求直线PQ的方程.

Answer 1

设P(x1,y1),Q(x2,y2)

则：(x1+x2)/2=2,(y1+y2)/2=1

x1^2/16+y1^2/9=1,x2^2/16+y2^2/9=1

所以

(x1^2-x2^2)/16+(y1^2-y2^2)/9=0

(y1+y2)(y1-y2)/9=-(x1-x2)(x1+x2)/16

(y1-y2)/(x1-x2)=-9(x1+x2)/16(y1+y2)=-36/32=-9/8

所以,PQ斜率为：-9/8

PQ直线方程为：

y-1=-9/8*(x-2)

即：8y+9x=26