如图，椭圆\(C_{1}： \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}}=1(a > b > 0)\)的离心率为\( \dfrac { \sqrt {2}}{2}\)，\(x\)轴被曲线\(C_{2}：y=x^{2}-b\)截得的线段长等于\(C_{1}\)的短轴长\(.C_{2}\)与\(y\)轴的交点为\(M\)，过坐标原点\(O\)的直线\(l\)与\(C_{2}\)相交于点\(A\)、\(B\)，直线\(MA\)，\(MB\)分别与\(C_{1}\)相交于点\(D\)、\(E\)．
\((1)\)求\(C_{1}\)、\(C_{2}\)的方程；
\((2)\)求证：\(MA⊥MB\)．
\((3)\)记\(\triangle MAB\)，\(\triangle MDE\)的面积分别为\(S_{1}\)、\(S_{2}\)，若\( \dfrac {S_{1}}{S_{2}}=λ\)，求\(λ\)的取值范围．