略

设 $P(x_0,y_0),M(x_1,y_1),N(x_2,y_2)$，则椭圆过点 $M,N$ 的切线方程分别为$$\dfrac{x_1x}{25}+\dfrac{y_1y}{9}=1 , \dfrac{x_2x}{25}+\dfrac{y_2y}{9}=1,$$因为两切线都过点 $P$，则有$$\dfrac{x_1x_0}{25}+\dfrac{y_1y_0}{9}=1 , \dfrac{x_2x_0}{25}+\dfrac{y_2y_0}{9}=1,$$这表明 $M,N$ 均在直线$$\dfrac{x_0x}{25}+\dfrac{y_0y}{9}=1\qquad\cdots\cdots\text{ ① }$$上，由两点决定一条直线知，式 $\text{ ① }$ 就是直线 $MN$ 的方程，其中 $(x_0,y_0)$ 满足直线 $l$ 的方程．
当点 $P$ 在直线 $l$ 上运动时，可理解为 $x_0$ 取遍一切实数，相应的 $y_0$ 为$$y_0=\dfrac57x_0-10,$$代入 $\text{ ① }$ 式消去 $y_0$，得$$\dfrac{x_0}{25}x+\dfrac{5x_0-70}{63}-1=0\qquad\cdots\cdots\text{ ② }$$$\text{ ② }$ 式对一切 $x_0\in\mathbb R$ 恒成立，变形可得$$x_0\left(\dfrac{x}{25}+\dfrac{5y}{63}\right)-\left(\dfrac{10y}{9}+1\right)=0,$$对一切 $x_0\in\mathbb R$ 恒成立，故有$$\begin{cases}\dfrac{x}{25}+\dfrac{5y}{63}=0,\\ \dfrac{10y}{9}+1=0.\end{cases}$$由此解得直线 $MN$ 恒过定点 $Q\left(\dfrac{25}{14},-\dfrac{9}{10}\right)$．

略

当 $MN\parallel l$ 时，由 $\text{ ② }$ 式知，$$\dfrac{\frac{x_0}{25}}{5}=\dfrac{\frac{5x_0-70}{63}}{-7}\ne\dfrac{-1}{-70},$$解得 $x_0=\dfrac{4375}{533}$．代入 $\text{ ② }$ 式，得此时 $MN$ 的方程为$$5x-7y-\dfrac{533}{35}=0,\qquad\cdots\cdots\text{ ③ }$$将此方程与椭圆方程联立，消去 $y$ 得$$\dfrac{533}{25}x^2-\dfrac{533}{7}x-\dfrac{128068}{1225}=0,$$由此可得，此时 $MN$ 截椭圆所得弦的中点横坐标恰好为点 $Q\left(\dfrac{25}{14},-\dfrac{9}{10}\right)$ 的横坐标，即$$x=\dfrac{x_1+x_2}{2}=\dfrac{25}{14}.$$代入 $\text{ ③ }$ 式可得弦中点纵坐标恰好为点 $Q\left(\dfrac{25}{14},-\dfrac{9}{10}\right)$ 的纵坐标，即$$y=\dfrac57\cdot\dfrac{25}{14}-\dfrac{533}{7\cdot35}=-\dfrac{9}{10}.$$这就是说，点 $Q\left(\dfrac{25}{14},-\dfrac{9}{10}\right)$ 平分线段 $MN$．