已知椭圆$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$（$a > b > 0$），${F_1}$、${F_2}$为其左右焦点，$P$为椭圆$C$上任意一点，$I$为$\triangle P{F_1}{F_2}$内切圆圆心，点$G$满足$\overrightarrow {P{F_1}}+ \overrightarrow {P{F_2}}= 3\overrightarrow {PG} $且$\overrightarrow {GI}= \lambda \overrightarrow {{F_1}{F_2}} $（$\lambda\in {\mathbb {R}}$且$\lambda\ne 0$），则椭圆的离心率是________．
正确答案是$\dfrac{1}{2}$．
分析与解 如图，因为$\overrightarrow {GI}= \lambda \overrightarrow {{F_1}{F_2}} $，所以${y_G} = {y_I}$．
而$G$是重心，所以$${y_G} = \dfrac{1}{3}{y_P} = r,$$而内切圆半径$$r = \dfrac{{2{S_{\triangle P{F_1}{F_2}}}}}{{P{F_1} + P{F_2} + {F_1}{F_2}}}= \dfrac{{2 \cdot \frac{1}{2} \cdot {F_1}{F_2} \cdot {y_P}}}{{2a + 2c}} = \dfrac{c}{{a + c}} \cdot {y_P},$$因此$\dfrac{1}{3}{y_P} = \dfrac{c}{{a + c}} \cdot {y_P}$，所以$\dfrac{c}{a} = \dfrac{1}{2}$．