Question 177, as shown in the figure, in the Cartesian coordinate system $\mathrm{xO} y$, $\mathrm{F}$ is a moving point on the positive half-axis of the $\mathrm{x}$-axis. A parabola $\mathrm{C}$ is constructed with $\mathrm{F}$ as the focus and $\mathrm{O}$ as the vertex. Let $\mathrm{P}$ be a point on $\mathrm{C}$ in the first quadrant, and $\mathrm{Q}$ be a point on the negative half-axis of the $\mathrm{x}$-axis such that $\mathrm{PQ}$ is a tangent to $\mathrm{C}$, and $|\mathrm{PQ}|=2$. Circles $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are both tangent to the line $\mathrm{OP}$ at point $\mathrm{P}$, and both are tangent to the $\mathrm{x}$-axis. Find the coordinates of point $\mathrm{F}$ such that the sum of the areas of circles $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is minimized.
Question 177, Solution: Let the equation of the parabola $C$ be $y^{2}=2 p x(p>0)$, then the coordinates of point $F$ are $\left(\frac{p}{2}, 0\right)$. Let point $P\left(2 p t^{2}, 2 p t\right)$, then $t>0$, and the slope of line $P Q$ is $k=\left.\frac{d y}{d x}\right|_{y=2 p t}=\left.\frac{p}{y}\right|_{y=2 p t}=\frac{1}{2 t}$. Therefore, $|\mathrm{PQ}|=2 \mathrm{pt} \cdot \sqrt{1+\frac{1}{\mathrm{k}^{2}}}=2 \mathrm{pt} \cdot \sqrt{1+4 \mathrm{t}^{2}}=2$, rearranging gives
$$
\mathrm{p}^{2} \mathrm{t}^{2}\left(1+4 \mathrm{t}^{2}\right)=1
$$

Also, $\mathrm{k}_{\mathrm{OP}}=\frac{1}{\mathrm{t}}, \angle \mathrm{O}_{2} \mathrm{OP}=\frac{\angle \mathrm{POF}}{2}$, so $\tan \angle \mathrm{POF}=\mathrm{k}_{\mathrm{OP}}=\frac{1}{\mathrm{t}}=\tan 2 \angle \mathrm{O}_{2} \mathrm{OP}=\frac{2 \tan \angle \mathrm{O}_{2} \mathrm{OP}}{1-\tan ^{2} \angle \mathrm{O}_{2} \mathrm{OP}}$
$$
\Rightarrow \tan \angle \mathrm{O}_{2} \mathrm{OP}=\sqrt{1+\mathrm{t}^{2}}-\mathrm{t}
$$

Let the centers of circles $C_{1}$ and $C_{2}$ be $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$, and their radii be $r_{1}$ and $r_{2}$, respectively, then
$$
\begin{array}{l}
r_{1}^{2}+r_{2}^{2}=O_{1} \mathrm{P}^{2}+\mathrm{O}_{2} \mathrm{P}^{2}=0 \mathrm{P}^{2}\left(\cot ^{2} \angle \mathrm{O}_{2} \mathrm{OP}+\tan ^{2} \angle \mathrm{O}_{2} \mathrm{OP}\right) \\
=O \mathrm{P}^{2}\left[\left(\frac{1}{\sqrt{1+t^{2}}-t}\right)^{2}+\left(\sqrt{1+\mathrm{t}^{2}}-\mathrm{t}\right)^{2}\right]=20 \mathrm{P}^{2} \cdot\left(1+2 \mathrm{t}^{2}\right) \\
=2\left[\left(2 \mathrm{pt}^{2}\right)^{2}+(2 p t)^{2}\right] \cdot\left(1+2 \mathrm{t}^{2}\right) \\
=8 \mathrm{p}^{2} \mathrm{t}^{2}\left(1+\mathrm{t}^{2}\right)\left(1+2 \mathrm{t}^{2}\right)
\end{array}
$$

To minimize the sum of the areas of circles $C_{1}$ and $C_{2}$, we need to minimize $r_{1}^{2}+r_{2}^{2}$. From (1) and (2), we know: $r_{1}^{2}+r_{2}^{2}=\frac{8\left(1+t^{2}\right)\left(1+2 t^{2}\right)}{1+4 t^{2}}=\left(4 t^{2}+1\right)+\frac{3}{4 t^{2}+1}+4$, then when $4 t^{2}+1=\sqrt{3}$, i.e., $\mathrm{t}^{2}=\frac{\sqrt{3}-1}{4}$, $r_{1}^{2}+r_{2}^{2}$ reaches its minimum value. At this point, $p^{2}=\frac{1}{\mathrm{t}^{2}\left(1+4 t^{2}\right)}=\frac{11}{\frac{(\sqrt{3}-1) \sqrt{3}}{4}}=\frac{4}{3-\sqrt{3}}$, thus $p=\frac{2}{\sqrt{3-\sqrt{3}}}$. Therefore, when the sum of the areas of circles $C_{1}$ and $C_{2}$ is minimized, the coordinates of point $F$ are $\left(\frac{1}{\sqrt{3-\sqrt{3}}}, 0\right)$.