Test

联立直线与椭圆方程：
\[\left\{\begin{array}{l} {y=kx+m} \\ {3x^{2} +4y^{2} =12} \end{array}\right. \Rightarrow \left(3+4k^{2} \right)x^{2} +8mkx+4\left(m^{2} -3\right)=0\]
\[{\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} x_{1} +x_{2} =-\frac{8mk}{4k^{2} +3} ,x_{1} x_{2} =\frac{4\left(m^{2} -3\right)}{4k^{2} +3} \]
\[{\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} y_{1} y_{2} =\left(kx_{1} +m\right)\left(kx_{2} +m\right)=k^{2} x_{1} x_{2} +mk\left(x_{1} +x_{2} \right)+m^{2} \]
      $=\frac{4k^{2} \left(m^{2} -3\right)}{4k^{2} +3} -\frac{8mk\cdot mk}{4k^{2} +3} +m^{2} =\frac{3m^{2} -12k^{2} }{4k^{2} +3} $，代入到①
\[\overrightarrow{DA}\cdot \overrightarrow{DB}=\frac{4\left(m^{2} -3\right)}{4k^{2} +3} +2\cdot \frac{8mk}{4k^{2} +3} +4+\frac{3m^{2} -12k^{2} }{4k^{2} +3} =0\]
\[{\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} \frac{4m^{2} -12+16mk+16k^{2} +12+3m^{2} -12k^{2} }{4k^{2} +3} =0\]
\[{\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} 7m^{2} +16mk+4k^{2} =0\Rightarrow \left(7m+2k\right)\left(m+2k\right)=0\]
${\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} m=-\frac{2}{7} k$或$m=-2k$

当$m=-\frac{2}{7} k$时，$l:y=kx-\frac{2}{7} k=k\left(x-\frac{2}{7} \right)$   ${\mkern1mu\raise1pt\hbox{.}\mkern1mu\raise4pt\hbox{.}\mkern1mu\raise1pt\hbox{.}\mkern1mu} l$恒过$\left(\frac{2}{7} ,0\right)$