内心向量式、焦半径、焦点弦长

设动点\(M(x,y)\)与定点\(F_2(\sqrt{2},0)\)的距离和它到定直线\(l:x=\dfrac{\sqrt{2}}{2}\)的距离比为\(\sqrt{2}\)，记点\(M\)的轨迹为\(C\)

(1)求\(C\)的方程

(2)设\(F_1(-\sqrt{2},0)\)，过点\(F_2\)的直线与\(C\)的右支相交于\(A,B\)两点，\(I\)是\(\triangle F_1AB\)内的一点，且满足\(|F_1B|\cdot\overrightarrow{IA}+|BA|\cdot\overrightarrow{IF_1}+|AF_1|\cdot\overrightarrow{IB}=\boldsymbol{0}\)，判断点\(I\)在定直线上.

解
(1)设\(M(x,y)\)，则有\((x-\sqrt{2})^2+y^2=2\left(x-\dfrac{1}{\sqrt{2}}\right)^2\)

整理有\(C:x^2-y^2=1\)

(2)设\(I(x_0,y_0),A(x_1,y_1),B(x_2,y_2)\)，\(AB:y=k(x-\sqrt{2})\)

联立\(\begin{cases} x^2-y^2=1\\ y=k(x-\sqrt{2}) \end{cases}\)有\((1-k^2)x^2+2\sqrt{2}k^2x-2k^2-1=0\)

即\(x_1+x_2=\dfrac{2\sqrt{2}k^2}{k^2-1},x_1x_2=\dfrac{2k^2+1}{k^2-1}\)

由题

\[|F_1b|(x_1-x_0,y_1-y_0)+|BA|(-\sqrt{2}-x_1,-y)+|AF_2|(x_2-x_0,y_2-y_0)=(0,0) \]

得$$x=\dfrac{|F_1B|x_1+|F_1A|x_2-\sqrt{2}|AB|}{|F_1B|+|AB|+|AF_1|}$$

由焦半径公式，得

\(|AF_1|=ex_1+a=\sqrt{2}+1\)

\(|BF_2|=ex_2+a=\sqrt{2}x_2+1\)

\(|AF_1|=ex_1-a=\sqrt2-a\)

\(|BF_2|=ex_2-a=\sqrt{2}x_2-a\)

\(|AB|=|AF_2|+|F_2B|=\sqrt{2}(x_1+x_2)-2\)

全部代回有

\(x=\dfrac{x_1+\sqrt{2}x_1x_2+x_2+\sqrt{2}x_1x_2-2(x_1+x_2)+2\sqrt{2}}{2+\sqrt{2}(x_1+x_2)+\sqrt{2}(x_1+x_2)-2}\)
\(=\dfrac{2\sqrt{2}x_1x_2-(x_1+x_2)+2\sqrt{2}}{2\sqrt{2}(x_1+x_2)}\)
\(=\dfrac{2\sqrt{2}(2k^2+1)-2\sqrt{2}k^2+2\sqrt{2}(k^2-1)}{8k^2}\)
\(=\dfrac{4\sqrt{2}k^2}{8k^2}\)
\(=\dfrac{\sqrt{2}}{2}\)