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以抛物线$y=\frac{x^2}{2p}$为例讨论,求导数得到切线斜率
$y'=\frac{x}{p}$.
对于抛物线上一点$(x_0,y_0)$($y_0=\frac{x_0^2}{2p}$)和抛物线外一点$(x_1,y_1)$,两点连线的斜率为
$k_1=\frac{y_1-y_0}{x_1-x_0}$,
$(x_0,y_0)$点切线的斜率为
$k_0=\frac{x_0}{p}$
要使两点连线垂直于切线,$k_0 k_1=-1$,即
$\frac{x_0(y_1-y_0)}{p(x_1-x_0)}=-1$
带入$y_0$得
$\frac{x_0(y_1-x_0^2/2p)}{p(x_1-x_0)}=-1$
$\frac{x_0^3}{2p}-(y_1-p) x_0-px_1=0$
将此一元三次方程写为$x_0^3-2p(y_1-p)x_0-2p^2x_1=0$和标准形式
x^3+bx+c=0对比得,
$b=-2p(y_1-p)$,$c=-2p^2$,判别式
$\Delta=\frac{c^2}{4}+\frac{b^3}{27}$
若$\Delta>0$,则唯一的实根为
$x_0=\left(-\frac{c}{2}+\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}\right)^{1/3}$
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