复旦大学 2023年强基第8题

解析：设 $P=(x, y)$ ，则 $\displaystyle \overrightarrow{M P}=\left(x-4, y-\frac{9}{5}\right), \overrightarrow{N P}=\left(x+4, y+\frac{9}{5}\right)$ ，故条件 $\angle M P N=105^{\circ}$ 等价于 $\displaystyle \frac{\overrightarrow{M P} \cdot \overrightarrow{P P}}{|\overrightarrow{M P}| \cdot|\overrightarrow{N P}|}=\frac{x^{2}-16+y^{2} \frac{B 1}{25}}{\sqrt{(x-4)^{2}+\left(y-\frac{9}{5}\right)^{2}} \cdot \sqrt{(x+4)^{2}+\left(y+\frac{9}{5}\right)^{2}}}=\cos 105^{\circ}$ ．由于 $\displaystyle \cos 105^{\circ}=-\sin 15^{\circ}=-\frac{\sqrt{6}-\sqrt{2}}{4}\lt 0$ ，则上述条件可化为 $\displaystyle \left\{\begin{array}{l}x^{2}-16+y^{2}-\frac{81}{25}\lt 0 \\ \frac{1}{\sqrt{1+\frac{\left(\frac{18}{5} x-8 y\right)^{2}}{\left(x^{2}-16+y^{2}-\frac{81}{25}\right)^{2}}}}=\frac{\sqrt{6}-\sqrt{2}}{4}\end{array}\right.$ ．由第二个方程可解出 $\displaystyle \pm(2+\sqrt{3})=\frac{\frac{18}{5} x-8 y}{x^{2}-16+y^{2} \frac{81}{25}}$ ，则上式可分为以下两种情形： （1）$\displaystyle \left\{\begin{array}{c}\frac{18}{5} x-8 y\lt 0 \\ \left(x-\frac{9}{5}(2-\sqrt{3})\right)^{2}+(y+4(2-\sqrt{3}))^{2}=481\left(\frac{\sqrt{6}-\sqrt{2}}{5}\right)^{2} ;\end{array}\right.$ （2）$\displaystyle \left\{\begin{array}{c}\frac{18}{5} x-8 y\gt 0 \\ \left(x+\frac{9}{5}(2-\sqrt{3})\right)^{2}+(y-4(2-\sqrt{3}))^{2}=481\left(\frac{\sqrt{6}-\sqrt{2}}{5}\right)^{2} .\end{array}\right.$ 在两种情形中，$(x, y)$ 的轨迹都是一个半圆，它的两个端点恰为 $M, N$ ，与椭圆交于一点．因此椭圆上满足条件的点恰有两个。