9. Given a point $P$ (not the origin) on the parabola $4 y=x^{2}$, the tangent line at $P$ intersects the $x$-axis and $y$-axis at points $Q$ and $R$ respectively, and $F$ is the focus. If a point $Q$ on the parabola satisfies $\overrightarrow{P F}=\mu \overrightarrow{F Q}$, find the minimum area of $\triangle Q P R$ and write the equation of the tangent line $P R$ at this time.
9. Solution: Let $P\left(t, \frac{t^{2}}{4}\right)(t \neq 0)$, then $l_{P R}: y-\frac{t^{2}}{4}=\frac{t}{2}(x-t)$, which gives $R\left(0,-\frac{t^{2}}{4}\right)$; from $l_{P Q}: y-1=\frac{\frac{t^{2}}{4}-1}{t} x$ and the parabola $4 y=x^{2}$, we solve to get $Q\left(-\frac{4}{t}, \frac{4}{t^{2}}\right)$.
Therefore: $S_{\triangle P Q R}=\frac{1}{2}|R F|\left|x_{P}-x_{Q}\right|=\frac{1}{2}\left|1+\frac{t^{2}}{4}\right|\left|t+\frac{4}{t}\right|=\frac{1}{2}\left|\frac{t^{3}}{4}+2 t+\frac{4}{t}\right|$.
First, we examine the minimum value of the function $f(t)=\frac{1}{2}\left(\frac{t^{3}}{4}+2 t+\frac{4}{t}\right)$ when $t>0$.
Let $f^{\prime}(t)=\frac{1}{2}\left(\frac{3 t^{2}}{4}+2-\frac{4}{t^{2}}\right)=0$ to get $t=\frac{2 \sqrt{3}}{3}$.
When $t \in\left(0, \frac{2 \sqrt{3}}{3}\right)$, $f^{\prime}(t)0$;
So $f(t)$ attains its minimum value $f\left(\frac{2 \sqrt{3}}{3}\right)=\frac{16 \sqrt{3}}{9}$ when and only when $t=\frac{2 \sqrt{3}}{3}$.
Since $|f(t)|=\frac{1}{2}\left|\frac{t^{3}}{4}+2 t+\frac{4}{t}\right|$ is an even function of $t$, it also attains its minimum value $\frac{16 \sqrt{3}}{9}$ when $t=-\frac{2 \sqrt{3}}{3}$.
Therefore, the minimum area of $\triangle P Q R$ is $\frac{16 \sqrt{3}}{9}$, and at this time, the equation of the tangent line $P R$ is $y=\frac{\sqrt{3}}{3} x-\frac{1}{3}$ or $y=-\frac{\sqrt{3}}{3} x-\frac{1}{3}$.