已知F1,F2为双曲线C;x²-y²=2的左右焦点,点P在C上,亅PF1亅=2亅PF2亅,则COS角F1PF2=?
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![已知F1,F2为双曲线C;x²-y²=2的左右焦点,点P在C上,亅PF1亅=2亅PF2亅,则COS角F1PF2=?](/uploads/image/z/10195626-66-6.jpg?t=%E5%B7%B2%E7%9F%A5F1%2CF2%E4%B8%BA%E5%8F%8C%E6%9B%B2%E7%BA%BFC%EF%BC%9Bx%26%23178%3B-y%26%23178%3B%3D2%E7%9A%84%E5%B7%A6%E5%8F%B3%E7%84%A6%E7%82%B9%2C%E7%82%B9P%E5%9C%A8C%E4%B8%8A%2C%E4%BA%85PF1%E4%BA%85%3D2%E4%BA%85PF2%E4%BA%85%2C%E5%88%99COS%E8%A7%92F1PF2%3D%3F)
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已知F1,F2为双曲线C;x²-y²=2的左右焦点,点P在C上,亅PF1亅=2亅PF2亅,则COS角F1PF2=?
已知F1,F2为双曲线C;x²-y²=2的左右焦点,点P在C上,亅PF1亅=2亅PF2亅,则COS角F1PF2=?
已知F1,F2为双曲线C;x²-y²=2的左右焦点,点P在C上,亅PF1亅=2亅PF2亅,则COS角F1PF2=?
标准方程为:x²/2-y²/2=1
|PF1|=2|PF2|
|PF1|-|PF2|=2a=|PF2|=2√2
则|PF1|=4√2
F1F2=2c=4
由余弦定理:cos∠F1PF2
=(|PF1|²+|PF2|²-F1F2²)/2|PF1||PF2|
=(8+32-16)/32
=3/4