P是椭圆x^2/9+y^2/5=1上的动点,M,N分别为左右焦点,若|PM|*|PN|=2/(1-cosMPN),求P点坐标

题目

P是椭圆x^2/9+y^2/5=1上的动点,M,N分别为左右焦点,若|PM|*|PN|=2/(1-cosMPN),求P点坐标

答案

x^2/9+y^2/5=1,a=3,b=√5,c=2MN=2c=4PM+PN=2a=6PM*PN=2/(1-cosMPN)在△MPN中,由余弦定理,得MN^2=PM^2+PN^2-2PM*PN*cosMPN=(PM+PN)^2-2PM*PN(1+cosMPN)4^2=6^2-[2*2/(1-cosMPN)]*(1+cosMPN)cosMPN=2/3>0sinMPN=√5/3P...