已知椭圆x^2/a^2+y^2/b^2=1(a＞b＞0)上一点P(3,4),若PF⊥PE,试求椭圆方程

Question

已知椭圆x^2/a^2+y^2/b^2=1(a＞b＞0)上一点P(3,4),若PF⊥PE,试求椭圆方程

Answer 1

用矢量,内积为0
根据条件就可以了
P,Q的坐标用(x1,-x1-1),(x2,-x2-1)
则x1x2+x1x2+x1+x2+1=0.(*)
a^2=2c^2,b^2=c^2
x1^2/2c^2+y1^2/c^2=1.(1)
x2^2/2c^2+y2^2/c^2=1.(2)
(2)-(1)
(x1^2-x2^2)/2c^2+(y1+y2)(y1-y2)/c^2=0
(x1^2-x2^2)/2c^2+(-x1-x2-2)(x2-x1)/c^2=0.(3)
由(*),(3)
((x1-x2)(1.5(x1+x2)+2))/c^2=0
即：(x1-x2)(1.5(x1+x2)+2)=0.(4)
由（*)和（4）
（0.5-3x1x2)(x1-x2)=0
显然x1!=x2
x1x2=1/6,x1+x2=-4/3
(1)+(2)
(x1^2+x2^2)/2c^2+(y1^2+y2^2)/c^2=2
((x1+x2)^2-2x1x2)/2c^2+((-x1-x2-2)^2-2(x1+1)(x2+1))/c^2=2
即：
(16/9-1/3)/2c^2+((4/3-2)^2-2(1/6-4/3+1))/c^2=2
即：
(13/18c^2+14/18c^2=2
c^2=3/4
a^2=3/2,b^2=3/4
椭圆x^2/(3/2)+y^2/(3/4)=1