设抛物线 \(\Gamma:y^2=2px(p>0)\)，直线 \(l:x=my+p\) 经过 \(T(p,0)\) 并且与 \(\Gamma\) 交于两点 \(A(x_1,y_1),B(x_2,y_2)\)

求证：\(\frac{1}{|AT|^2}+\frac{1}{|BT|^2}=\frac{1}{p^2}\)

法一

\[\begin{aligned} &\begin{cases} y^2=2px \\ x=my+p \\ \end{cases} \Rightarrow y^2=2pmy+2p^2 \Rightarrow \begin{cases} y_1+y_2=2pm \\ y_1 \cdot y_2=-2p^2 \end{cases} \\ \frac{1}{|AT|^2}+\frac{1}{|BT|^2} =&\frac{1}{(x_1-p)^2+y_1^2}+\frac{1}{(x_2-p)^2+y_2^2}=\frac{1}{m^2+1}\left( \frac{1}{y_1^2}+\frac{1}{y_2^2} \right) \\ =&\frac{1}{m^2+1}\frac{(y_1+y_2)^2-2y_1y_2}{y_1^2y_2^2}=\frac{1}{m^2+1}\frac{4p^2m^2+4p^2}{4p^4}=\frac{1}{p^2} \end{aligned} \]

法二

考虑到在直角三角形中，对于直角边 \(a,b\) 和斜边上的高 \(h\) 有：\(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{h^2}\)

将 \(B\) 逆时针旋转到 \(l_{\perp}\) 上得到 \(B'(x_3,y_3)\)，于是有：

\[\begin{bmatrix} x_3 \\ y_3 \end{bmatrix}= \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} x_2-p \\ y_2 \end{bmatrix}+ \begin{bmatrix} p \\ 0 \end{bmatrix} = \begin{bmatrix} p-y_2 \\ x_2-p \end{bmatrix} \]

于是有 \(l_{AB'}:y=\frac{x_2-p-y_1}{p-y_2-x_1}(x-x_1)+y_1\)，接下来只需要验证 \(\Delta ATB'\) 中有 \(h=p\) 即可

\[\begin{aligned} h&=\frac{|(x_2-p-y_1)(p-x_1)+(p-y_2-x_1)y_1|}{\sqrt{(p-y_2-x_1)^2+(x_2-p-y_1)^2}} \\ &\xlongequal[x_2-p=my_2]{x_1-p=my_1}\frac{|(my_2-y_1)(-my_1)+(-y_2-my_1)y_1|}{\sqrt{(1+m^2)(y_1^2+y_2^2)}} \\ &=\frac{|(1+m^2)y_1y_2|}{\sqrt{(1+m^2)(4p^2(1+m^2))}}=\frac{2p^2}{2p}=p \end{aligned} \]