如图,在平面直角坐标系\(xOy\)中,椭圆\( \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1(a > b > 0)\)的左、右焦点分别为\(F_{1}(-c,0)\),\(F_{2}(c,0).\)已知\((1,e)\)和\((e, \dfrac { \sqrt {3}}{2})\)都在椭圆上,其中\(e\)为椭圆的离心率.
\((1)\)求椭圆的方程;
\((2)\)设\(A\),\(B\)是椭圆上位于\(x\)轴上方的两点,且直线\(AF_{1}\)与直线\(BF_{2}\)平行,\(AF_{2}\)与\(BF_{1}\)交于点\(P\).
\((i)\)若\(AF_{1}-BF_{2}= \dfrac { \sqrt {6}}{2}\),求直线\(AF_{1}\)的斜率;
\((ii)\)求证:\(PF_{1}+PF_{2}\)是定值.