Question

椭圆x^2/a^2+y^2/b^2若点P是椭圆上任意一点，

椭圆x^2/a^2+y^2/b^2=1,，若点P是椭圆上任意一点，过原点直线L与椭圆交于M N两点，记直线PM ,PN的斜率分别为KPM，KPN，当KPM*KPN= -1/4 ，求椭圆方程

Answer 1

P(x0,y0),M(x1,y1),N(x2,y2)
过原点x2=-x1,y2=-y1
N(-x1,-y1)
KPM*KPN=(y1-y0)/(x1-x0) * (-y1-y0)/(-x1-x0)
=(y1-y0)/(x1-x0) * (y1+y0)/(x1+x0)
=(y1^2-y0^2)/(x1^2-x0^2)
x1^2/a^2+y1^2/b^2=1
x0^2/a^2+y0^2/b^2=1
两式相减，(x1^2-x0^2)/a^2+(y1^2-y0^2)/b^2=0
整理得(y1^2-y0^2)/(x1^2-x0^2)=-b^2/a^2
-b^2/a^2=-1/4
b^2/a^2=1/4
由于缺条件，所以椭圆方程x^2/(4b^2)+y^2/b^2=1