Question

椭圆x^2/9+y^2/4=1的焦点为F1、F2,点P为其上动点,当∠F1PF2为钝角时,求点P的横坐标的取值范围

Answer 1

c^2=a^2-b^2=5 F1(-根号5,0) F2(根号5,0)
设P(x0,y0)，当∠F1PF2为钝角时，有(x0+根号5)(x0-根号5)+y0^2<0
即x0^2+y0^2-5<0 x0^2+y0^2<5
x0^2/9+y0^2/4=1 y0^2=4-4x0^2/9
5x0^2/9+4<5 x0^2<9/5 -3根号5/5<x0<3根号5/5
y0^2>4-4(9/5)/9=16/5 -2<=y0<-4根号5/5或4根号5/5<y0<=2

Answer 2

c^2=a^2-b^2=5 F1(-根号5,0) F2(根号5,0)
设P(x0,y0)，当∠F1PF2为钝角时，有(x0+根号5)(x0-根号5)+y0^2<0
即x0^2+y0^2-5<0 x0^2+y0^2<5
x0^2/9+y0^2/4=1 y0^2=4-4x0^2/9
5x0^2/9+4<5 x0^2<9/5 -3根号5/5<x0<3根号5/5
y0^2>4-4(9/5)/9=16/5 -2<=y0<-4根号5/5或4根号5/5<y0<=2