25~26高三下·东北三省三校二模·第8题

$\boldsymbol{C}$

$\because \overrightarrow{MF_1}=2\overrightarrow{MH}$，故点$H$位于线段$MF_1$的中点，

设$M(x_0,y_0)$，$F_1(-c,0)$，$F_2(c,0)$，则$H\left(\dfrac{x_0-c}{2},\dfrac{y_0}{2}\right)$，

$\overrightarrow{MF_1}=(-c-x_0,-y_0)$，$\overrightarrow{HF_2}=\left(\dfrac{3c-x_0}{2},-\dfrac{y_0}{2}\right)$，

又$\overrightarrow{MF_1}\cdot\overrightarrow{HF_2}=0$，$\therefore (-c-x_0)\left(\dfrac{3c-x_0}{2}\right)+(-y_0)\left(-\dfrac{y_0}{2}\right)=0$，化简得
$x_0^2-2cx_0+y_0^2=3c^2.$

又$M$为椭圆上一点，$\therefore \dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=1$，

联立得$x_0^2-2cx_0+b^2\left(1-\dfrac{x_0^2}{a^2}\right)=3c^2$，

即$\dfrac{c^2}{a^2}x_0^2-2cx_0+(b^2-3c^2)=0$，此方程有解，

则$(-2c)^2-4\times\dfrac{c^2}{a^2}\times(b^2-3c^2)\geq0$，

化简得$a^2\geq b^2-3c^2$，即$a^2\geq a^2-4c^2$恒成立，

方程$\dfrac{c^2}{a^2}x_0^2-2cx_0+(b^2-3c^2)=0$$\Rightarrow c^2x_0^2-2a^2cx_0+a^4-4a^2c^2=0$的两个根为$\dfrac{a^2+2ac}{c}$和$\dfrac{a^2-2ac}{c}$，

又$x_0\in[-a,a]$，$\dfrac{a^2+2ac}{c}=\dfrac{a^2}{c}+2a>a$，

$\therefore -a\leq\dfrac{a^2-2ac}{c}\leq a$，解得$e\geq\dfrac{1}{3}$，

则$C$的离心率的取值范围是$[\frac{1}{3},1)$