椭圆的两个顶点A(-1,0),B(1,0),过其焦点F(0,1)的直线L与椭圆交于两点CD,并与X轴交与点P当

椭圆的两个顶点A(-1,0),B(1,0),过其焦点F(0,1)的直线l与椭圆交于两点CD,并与X轴交与点P当绝对值CD等于二分之三倍根号二时,求直线l的方程... 椭圆的两个顶点A(-1,0),B(1,0),过其焦点F(0,1)的直线l与椭圆交于两点CD,并与X轴交与点P当绝对值CD等于二分之三倍根号二时,求直线l的方程 展开
littlepigus
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采用椭圆标准方程
椭圆的两个顶点A(-1,0),B(1,0)=>a^2=1
焦点F(0,1)=>b^2-a^2=1 =>b^2=2
椭圆方程:x^2+y^2/2=1
直线L过(0,1), 可设直线l的方程为
(y-1)=kx =>y=kx+1
代入椭圆方程
x^2+(kx+1)^2/2=1
=>(1+k^2/2)x^2+kx-1/2=0
=>(2+k^2)x^2+2kx-1=0 (1)
设CD两点坐标为(x1,y1)(x2,y2),x1,x2为方程(1)的两个根。
|CD|=√[(x2-x1)^2+(y2-y1)^2]=√(1+k^2)|x2-x1|=√(1+k^2)×√[(x1+x2)^2-4x1x2]
=√(1+k^2)×√[(2k/(2+k^2))^2+4/(2+k^2)]=3√2/2
=>(1+k^2)×(4k^2+8+4k^2)=(9/2)×(2+k^2)^2
=>[(1+k^2)/(2+k^2)]^2=9/16
=>k^2=2
=>k=±√2
直线方程为y=±√2x+1
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直线方程为y=±√2x+1
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