6.
如图,\(\triangle ABC\)中,点\(A(-1,0)\),\(B(1,0).\)圆\(I\)是\(\triangle ABC\)的内切圆,且\(CI\)延长线交\(AB\)与点\(D\),若\(\overrightarrow{CI}=2\overrightarrow{ID}\).
\((1)\)求点\(C\)的轨迹\(Ω\)的方程;
\((2)\)若椭圆\(\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a > b > 0)\)上点\((x_{0},y_{0})\)处的切线方程是\(\dfrac{{{x}_{0}}x}{{{a}^{2}}}+\dfrac{{{y}_{0}}y}{{{b}^{2}}}=1\),过直线\(l\):\(x=4\)上一点\(M\)引\(Ω\)的两条切线,切点分别是\(P\)、\(Q\),求证:直线\(PQ\)恒过定点\(N\).