已知F1,F2是椭圆(X^2)/9+(Y^2)/4=1的两个焦点,点P在椭圆上,如果△PF1F2是直角三角形,求点P的坐标
此题一共有八种情况.我会求四种P的坐标.若以P为直角,P的四个坐标为什么?
人气:318 ℃ 时间:2019-08-20 04:30:20
解答
若以P为直角,
PF1^2+PF2^2=F1F2^2,
(x-√5)^2+y^2+(x+√5)^2+y^2=(2√5)^2,
x^2+y^2=5,
与x^2/9+y^2/4=1联立,解得P的四个坐标
(3√5/5,4√5/5),(3√5/5,-4√5/5),(-3√5/5,4√5/5),(-3√5/5,-4√5/5).
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