一个有关抛物线上定点的二级结论

抛物线上的定点问题1例,及与之相关的一个二级结论。


1. 已知抛物线y^2 = 2px \, (p>0),过点M(5,-2)的动直线l交抛物线于AB两点,线段AB的中点为N,当直线l的斜率为-1时,点M恰好为线段AB的中点。

(1) 求抛物线的方程;

(2) 抛物线上是否存在一个定点P,使得|PN| = \dfrac12 |AB|,若存在,求出点P坐标;若不存在,请说明理由。


分析与解    (1) y^2 = 4x,过程略。

(2) 设直线lx = m(y+2) + 5A(x_1,y_1)B(x_2,y_2)。假设存在点P(4t^2,4t)

\begin{cases} y^2 = 4x \\ x = m(y+2) + 5 \end{cases},得y^2 - 4my - 8m - 20 = 0\therefore y_1 + y_2 = 4my_1y_2 = -8m-20

x_1+x_2 = m(y_1 + y_2) + 4m + 10 = 4m^2 + 4m + 10x_1x_2 = \dfrac{(y_1y_2)^2}{16} = (2m+5)^2


|PN| = \dfrac12 |AB|等价于PA \bot PB,即\overrightarrow{PA} \cdot \overrightarrow{PB} = 0

\overrightarrow{PA} = (x_1 - 4t^2, y_1 - 4t)\overrightarrow{PB} = (x_2 - 4t^2, y_2 - 4t)。所以

\overrightarrow{PA} \cdot \overrightarrow{PB} = (x_1 - 4t^2)(x_2 - 4t^2) + (y_1 - 4t)(y_2 - 4t)

= x_1x_2 - 4t^2(x_1+x_2) + 16t^4 + y_1y_2 - 4t(y_1 + y_2) + 16t^2

= (4-16t^2)m^2 + (12-16t-16t^2)m + 16t^4 - 24t^2 + 5 = 0

要使该式对变化的m恒成立,需\begin{cases} 4-16t^2 = 0 \\ 12-16t-16t^2=0 \\ 16t^4 - 24t^2 + 5=0 \end{cases}

t = \dfrac12满足条件,此时P(1,2)

故存在点P使满足题意要求,P的坐标为(1,2)


拓展    抛物线中有一个与本题相关的二级结论:点P(x_0,y_0)为抛物线y^2 = 2px \, (p>0)上的一个点,AB为该抛物线上的动点,且满足PA \bot PB,则直线AB过定点Q(x_0 + 2p, -y_0)


证明    设Q(s,t)ABx = m(y-t) + sA(x_1,y_1)B(x_2,y_2)

联立\begin{cases} y^2 = 2px \\ x = my - mt + s \end{cases},得y^2 - 2pmy + 2pmt - 2ps = 0

所以y_1+y_2 = 2pmy_1y_2 = 2pmt - 2ps

x_1+x_2 = 2pm^2 - 2mt + 2sx_1x_2 = (mt-s)^2

\overrightarrow{PA} \cdot \overrightarrow{PB} = x_1x_2 - x_0(x_1+x_2) + x_0^2 + y_1y_2 - y_0(y_1+y_2) + y_0^2

= (t^2 - 2px_0)m^2 + (2tx_0 - 2ts + 2pt -2py_0)m + s^2 - 2sx_0 + x_0^2 - 2ps + y_0^2 = 0

解得\begin{cases}s = x_0 \\ t = y_0 \end{cases}(舍),或\begin{cases}s = x_0+2p \\ t = -y_0 \end{cases}

Q(x_0 + 2p, -y_0)

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