Question 180, In the Cartesian coordinate system, the hyperbola $\Gamma: \frac{x^{2}}{3}-y^{2}=1$. For any point $P$ in the plane not on $\Gamma$, let $\Omega_{p}$ be the set of all lines passing through point $P$ and intersecting $\Gamma$ at two points. For any line $l \in \Omega_{p}$, let $M, N$ be the two intersection points of $l$ and $\Gamma$, and define $f_{P}(l)=|P M| \cdot|P N|$. If there exists a line $l_{0} \in \Omega_{p}$ such that: the two intersection points of $l_{0}$ and $\Gamma$ are on opposite sides of the $y$-axis, and for any line $l \in \Omega_{p}, l \neq l_{0}$, we have $f_{P}(l)>f_{P}\left(l_{0}\right)$, then $P$ is called a “good point”. Find the area of the region formed by all good points.
Question 180, Answer: Let $P(a, b)$ be a good point. For any $l \in \Omega_{p}$, let $1: y=k x+c$, where $c=b-ka$. Solving the system of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{x^{2}}{3}-y^{2}=1 \\
y=k x+c
\end{array} \Rightarrow \frac{x^{2}}{3}-(k x+c)^{2}=1\right. \\
\Rightarrow\left(1-3 k^{2}\right) x^{2}-6 k c x-3\left(c^{2}+1\right)=0
\end{array}
$$

Since $I$ intersects $\Gamma$ at two points $M$ and $N$, we have $\Delta=(6 k c)^{2}+4\left(1-3 k^{2}\right) \cdot 3\left(c^{2}+1\right)=12\left(c^{2}+1-3 k^{2}\right)$ $>0$, which is equivalent to
$$
c^{2}>3 k^{2}-1 \Leftrightarrow (b-ka)^{2}>3 k^{2}-1
$$

According to Vieta's formulas, we have:
$$
\left\{\begin{array}{l}
x_{M}+x_{N}=\frac{6 k c}{1-3 k^{2}} \\
x_{M} \cdot x_{N}=\frac{-3\left(c^{2}+1\right)}{1-3 k^{2}}
\end{array}\right.
$$

Thus, according to the definition, we have:
$$
\begin{array}{l}
f_{P}(1)=\sqrt{1+k^{2}}\left|x_{P}-x_{M}\right| \cdot \sqrt{1+k^{2}}\left|x_{P}-x_{N}\right| \\
=\left(1+k^{2}\right)\left|a^{2}-a \cdot \frac{6 k c}{1-3 k^{2}}+\frac{-3\left(c^{2}+1\right)}{1-3 k^{2}}\right| \\
=\frac{\left(1+k^{2}\right)\left|a^{2}-3-3(a k+c)^{2}\right|}{\left|1-3 k^{2}\right|}=\frac{\left(1+k^{2}\right)\left|a^{2}-3-3 b^{2}\right|}{\left|1-3 k^{2}\right|}
\end{array}
$$

Moreover, $M$ and $N$ being on opposite sides of the $y$-axis is equivalent to $x_{M} \cdot x_{N}=\frac{-3\left(c^{2}+1\right)}{1-3 k^{2}} < 0 \Leftrightarrow 3 k^{2}-1 > 0$. When $M$ and $N$ are on the same side of the $y$-axis, it is equivalent to $x_{M} \cdot x_{N}=\frac{-3\left(c^{2}+1\right)}{1-3 k^{2}} > 0 \Leftrightarrow 3 k^{2}-1 > 0$.

When $M$ and $N$ are on opposite sides of the $y$-axis, $f_{p}(l)=\frac{\left(1+k^{2}\right)\left|a^{2}-3-3 b^{2}\right|}{1-3 k^{2}} \geq \left|a^{2}-3-3 b^{2}\right|$, and equality holds when $k=0$. Therefore, the minimum value of $f_{p}(l)$ is $f_{p}\left(l_{0}\right)=\left|a^{2}-3-3 b^{2}\right|$.

When $M$ and $N$ are on the same side of the $y$-axis, $f_{p}(l)=\frac{\left(1+k^{2}\right)\left|a^{2}-3-3 b^{2}\right|}{3 k^{2}-1}$. By the minimality of $f_{p}(16)=\left|a^{2}-3-3 b^{2}\right|$, this requires that for any $a, b, k$ satisfying (2) and $3 k^{2}-1 > 0$,
$$
\frac{\left(1+k^{2}\right)\left|a^{2}-3-3 b^{2}\right|}{3 k^{2}-1} > \left|a^{2}-3-3 b^{2}\right| \Leftrightarrow k^{2} < 1
$$

In other words, this condition is equivalent to: when $k^{2} \geq 1$, (2) does not hold.
On one hand, we take $k = \pm 1$, then according to the negation of (2), we have $(b \pm a)^{2} \leq 2 \Leftrightarrow |a| + |b| \leq \sqrt{2}$;
On the other hand, when $|a| + |b| \leq \sqrt{2}$, for any real number $k$ satisfying $k^{2} \geq 1$, we have
$$
(b-k a)^{2} \leq (|a k| + |b|)^{2} \leq (|a k| + |b k|)^{2} = (|a| + |b|)^{2} k^{2} \leq 2 k^{2} \leq 3 k^{2} - 1
$$

Thus, the negation of (2) holds.
Therefore, $P(a, b)$ is a good point if and only if $|a| + |b| \leq \sqrt{2}$, and the area of the region it forms is 4.