巧妙转化证垂直

已知$P$是定直线$y=n$($n<0$)上的一点，过$P$作抛物线$C:x^2=2py$($p>0$)的两条切线，设切点分别为$A(x_1,y_1)$，$B(x_2,y_2)$．
(1) 求证：$x_1x_2+y_1y_2$是定值；

(2) 设$\triangle PAB$的外接圆圆心为$M$，抛物线$C$的焦点为$F$，求证：$FM\perp FP$．

答案　(1) 定值为$2pn+n^2$；(2)略．

分析与解　(1) 设$P(m,n)$，则\[AB:mx=p(y+n),\]联立直线$AB$与抛物线方程，可得\[x^2-2mx+2pn=0,\]于是\[x_1+x_2=2m,x_1x_2=2pn.\]因此\[x_1x_2+y_1y_2=x_1x_2+\dfrac{(x_1x_2)^2}{4p^2}=2pn+n^2,\]为定值．

(2) 设$P(m,n)$点关于$F$点的对称点为$Q(-m,p-n)$，则问题等价于证明$P,A,B,Q$四点共圆．
设圆$M$的方程为\[x^2+y^2+Dx+Ey+F=0,\]则$(D,E,F)$满足方程组\[\begin{cases}
x_1^2+y_1^2+Dx_1+Ey_1+F=0,\\x_2^2+y_2^2+Dx_2+Ey_2+F=0,\\m^2+n^2+Dm+En+F=0,\end{cases}\]前两个方程相加可得\[x_1^2+x_2^2+y_1^2+y_2^2+(x_1+x_2)D+(y_1+y_2)E+2F=0,\]而\[\begin{aligned}&x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=4m^2-4pn,\\&y_1+y_2=\dfrac{x_1^2+x_2^2}{2p}=\dfrac{2m^2}p-2n,\\&y_1^2+y_2^2=\dfrac{\left(x_1^2+x_2^2\right)^2-2x_1^2x_2^2}{4p^2}=\dfrac{4m^4}{p^2}-\dfrac{4m^2n}p+2n^2,\end{aligned}\]代入可得\[4m^2-4pn+\dfrac{4m^4}{p^2}-\dfrac{8m^2n}p+2n^2+2mD+\left(\dfrac{2m^2}p-2n\right)E+2F=0,\]该方程与第三个方程的两倍相减，可得\[\left(\dfrac{m^2}p-2n\right)\left(E+\dfrac{2m^2}p+p\right)=0,\]于是解得\[E=-\dfrac{2m^2}p-p.\]前两个方程相减可得\[x_1^2-x_2^2+y_1^2-y_2^2+(x_1-x_2)D+(y_1-y_2)E=0,\]即\[x_1+x_2+\dfrac{(x_1+x_2)\left(x_1^2+x_2^2\right)}{4p^2}+D+\dfrac{x_1+x_2}{2p}E=0,\]也即\[2m+\dfrac{2m\left(4m^2-4pn\right)}{4p^2}+D+\dfrac{2m}{2p}E=0,\]将$E=-\dfrac{2m^2}p-p$代入可得\[D=\dfrac{2mn}p-m.\]欲证\[m^2+(p-n)^2+D(-m)+E(p-n)+F=0,\]只需要证明\[2np-p^2+2mD+(2n-p)E=0,\]将$D=\dfrac{2mn}p-m$，$E=-\dfrac{2m^2}p-p$代入检验即得原命题得证．

注　任何三点都不共线的四点$A(x_1,y_1)$，$B(x_2,y_2)$，$C(x_3,y_3)$，$D(x_4,y_4)$共圆的充分必要条件是\[\begin{vmatrix}
x_1^2+y_1^2 &x_1 &y_1 &1\\
x_2^2+y_2^2 &x_2 &y_2 &1\\
x_3^2+y_3^2 &x_3 &y_3 &1\\
x_4^2+y_4^2 &x_4 &y_4 &1\\
\end{vmatrix}=0.\]