Question

已知椭圆x^2/2+y^2=1,椭圆上有两点P.Q,O为原点,且有直线OP.OQ的斜率满足Kop*Koq=-1/2求线段

Answer 1

用尽可能少的参数解题.

设M(x,y),P(x+p,y+q),Q(x-p,y-q),依题意

(x+p)^2/2+(y+q)^2=1,①

(x-p)^2/2+(y-q)^2=1,②

(y+q)(y-q)/[(x+p)(x-p)]=-1/2,③

①+②,x^2+p^2+2y^2+2q^2=2,

p^2+2q^2=2-x^2-2y^2,

由③,x^2+2y^2=p^2+2q^2=2-x^2-2y^2,

∴x^2+2y^2=1,为所求.