椭圆x^2/9+y^2/4=1的焦点F1.F2,点P为其上的动点,当角F1PF2为钝角时,点P横坐标的取值范围为________?
人气:410 ℃ 时间:2019-11-12 14:47:48
解答
那么∠F1PF2的范围为(90,180)
:先求当∠=90时
设P(3sinθ,2cosθ)
由PF1⊥PF2
→2cos²θ/(9sinθ²-5)=-1
→sinθ=√5/5
→P(√5/5,cosθ)
当∠F1PF2=180,P(3,0)
→目标ε(√5/5,3)
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