已知抛物线x^2=2y,F是抛物线的焦点,过点F的直线L与抛物线相交于A、B两点,分别过A、B作抛物线L1、L2,记L1和L2相交于点M.1.证明L1⊥L22、求点M的轨迹方程已知抛物线x^2=2y,F是抛物线的焦点,过点F的

已知抛物线x^2=2y,F是抛物线的焦点,过点F的直线L与抛物线相交于A、B两点,分别过A、B作抛物线L1、L2,记L1和L2相交于点M.1.证明L1⊥L22、求点M的轨迹方程已知抛物线x^2=2y,F是抛物线的焦点,过点F的
已知抛物线x^2=2y,F是抛物线的焦点,过点F的直线L与抛物线相交于A、B两点,分别过A、B作抛物线L1、L2,记L1和L2相交于点M.
1.证明L1⊥L2
2、求点M的轨迹方程
已知抛物线x^2=2y,F是抛物线的焦点,过点F的直线L与抛物线相交于A、B两点,分别过A、B做抛物线的两条切线L1、L2记L1和L2相交于点M。第2问。。。主要是第2问。

已知抛物线x^2=2y,F是抛物线的焦点,过点F的直线L与抛物线相交于A、B两点,分别过A、B作抛物线L1、L2,记L1和L2相交于点M.1.证明L1⊥L22、求点M的轨迹方程已知抛物线x^2=2y,F是抛物线的焦点,过点F的
应该是分别过A、B作抛物线切线L1、L2吧
F（0,1/2）
直线L方程设为：y=kx+1/2
代入x2=2y中
x2-2kx-1=0
x1=√(1+k2)+k y1=k2+k√(1+k2)+1/2
x2=-√(1+k2)+k y2=k2-k√(1+k2)+1/2
A(√(1+k2)+k,k2+k√(1+k2)+1/2)
B(-√(1+k2)+k,k2-k√(1+k2)+1/2)
又y=x^2/2
y'=x
所以点A斜率Ka=√(1+k2)+k
点B斜率Kb=-√(1+k2)+k
Ka*Kb=-1
所以L1⊥L2

朋友，题目是不是抄错了哦，请检查一下！
特别是：分别过A、B作抛物线L1、L2，记L1和L2相交于点M。
证明：F（0,0.5）直线方程设为y=kx+0.5
y=kx+0.5 与x²=2y 联立方程组得x²-2kx-1=0 即 x1+x2=2k；x1·x2=-1
x²=2y变为 y=1/2x² 曲线的斜率为x，即L...

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朋友，题目是不是抄错了哦，请检查一下！
特别是：分别过A、B作抛物线L1、L2，记L1和L2相交于点M。
证明：F（0,0.5）直线方程设为y=kx+0.5
y=kx+0.5 与x²=2y 联立方程组得x²-2kx-1=0 即 x1+x2=2k；x1·x2=-1
x²=2y变为 y=1/2x² 曲线的斜率为x，即L1、L2的斜率分别为x1、x2
∵x1·x2=-1 ∴L1⊥L2 （注意：这里用了导数的思想）

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,so the shirt stays tucked in when you play
Ralph Lauren continues to be outfitting males inside the finest clothing for four decades. The signature polo shirt is one of the finest sport shirts co...

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,so the shirt stays tucked in when you play
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很容易的。
F（0，1/2）
直线L方程设为：y=kx+1/2
代入x2=2y中
x2-2kx-1=0
x1=√(1+k2)+k y1=k2+k√(1+k2)+1/2
x2=-√(1+k2)+k y2=k2-k√(1+k2)+1/2
A(√(1+k2)+k,k2+k√(1+k2)+1/2)
B(-√(1+...

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很容易的。
F（0，1/2）
直线L方程设为：y=kx+1/2
代入x2=2y中
x2-2kx-1=0
x1=√(1+k2)+k y1=k2+k√(1+k2)+1/2
x2=-√(1+k2)+k y2=k2-k√(1+k2)+1/2
A(√(1+k2)+k,k2+k√(1+k2)+1/2)
B(-√(1+k2)+k,k2-k√(1+k2)+1/2)
又y=x^2/2
y'=x
所以点A斜率Ka=√(1+k2)+k
点B斜率Kb=-√(1+k2)+k
Ka*Kb=-1
所以L1⊥L2

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