设 $P_1(x_1,y_1)$，$P_2(x_2,y_2)$，$M(x,y)$，且直线 $l$ 的倾斜角为 $\theta$，则$$\begin{cases} x_1=x+\dfrac{\sqrt2}{2}\cos\theta,\\ y_1=y+\dfrac{\sqrt2}{2}\sin\theta,\end{cases}$$且$$\begin{cases} x_2=x-\dfrac{\sqrt2}{2}\cos\theta,\\ y_2=y-\dfrac{\sqrt2}{2}\sin\theta, \end{cases}$$代入方程 $\dfrac{x^2}2+y^2=1$ 得$$\begin{cases} \mathrm{Eq1}:\left(x+\dfrac{\sqrt2}{2}\cos\theta\right)^2+2\left(y+\dfrac{\sqrt2}{2}\sin\theta\right)^2=2,\\
\mathrm{Eq2}:\left(x-\dfrac{\sqrt2}{2}\cos\theta\right)^2+2\left(y-\dfrac{\sqrt2}{2}\sin\theta\right)^2=2,
\end{cases}$$将 $ \mathrm{Eq1}+\mathrm{Eq2}$ 可得$$\mathrm{Eq3}:2x^2+4y^2+\sin^2\theta=3, $$将 $\mathrm{Eq1}-\mathrm{Eq2}$ 可得$$\tan\theta=-\dfrac{x}{2y}, $$所以$$\sin^2\theta=\dfrac{x^2}{x^2+4y^2},$$代入 $\mathrm{Eq3}$ 即得所求轨迹方程为$$2x^2+4y^2+\dfrac{x^2}{x^2+4y^2}=3.$$