9.
已知椭圆\(C:\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1(a > b > 0)\)的左、右焦点分别为\({{F}_{1}},{{F}_{2}}\),椭圆\(C\)过点\(P(1\ ,\ \dfrac{\sqrt{2}}{2})\),直线\(P{{F}_{1}}\)交\(y\)轴于\(Q\),且\(\overrightarrow{P{{F}_{2}}}=2\overrightarrow{QO}\),\(O\)为坐标原点.
\((1)\)求椭圆\(C\)的方程;
\((2)\)设\(M\)是椭圆\(C\)的上顶点,过点\(M\)分别作直线\(MA\ ,\ MB\)交椭圆\(C\)于\(A,B\)两点,设这两条直线的斜率分别为\({{k}_{1}}\ ,\ {{k}_{2}}\),且\({{k}_{1}}+{{k}_{2}}=2\),若直线\(AB\)斜率存在,求证:直线\(AB\)过定点.