Question 178: In the Cartesian coordinate system $x O y$, let a moving line $l$ be tangent to the parabola $\Gamma: y^{2}=4 x$, and intersect the hyperbola $\Omega: x^{2}-y^{2}=1$ at points $A$ and $B$ on the left and right branches, respectively. Find the minimum value of the area of $\triangle A O B$.
Question 178, Solution: Let $\mathrm{l}: \mathrm{y}=\mathrm{kx}+\mathrm{b}$. By symmetry, we can assume $\mathrm{k} \geq 0$.
Since 1 is tangent to $\Gamma$, solving the system of equations $\left\{\begin{array}{c}y=k x+b \\ y^{2}=4 x\end{array}\right.$, we get
$$
(k x+b)^{2}=4 x \Rightarrow k^{2} x^{2}+(2 k b-4) x+b^{2}=0
$$

The discriminant is $\Delta_{1}=(2 k b-4)^{2}-4 k^{2} b^{2}=0 \Rightarrow k b=1$, hence $b=\frac{1}{k}$.
Solving the system of equations $\left\{\begin{array}{l}y=k x+\frac{1}{k} \\ x^{2}-y^{2}=1\end{array}\right.$, we get
$$
\mathrm{x}^{2}-\left(k x+\frac{1}{k}\right)^{2}=1 \Rightarrow \mathrm{k}^{2}\left(1-\mathrm{k}^{2}\right) \mathrm{x}^{2}-2 \mathrm{k}^{2} \mathrm{x}-\left(\mathrm{k}^{2}+1\right)=0
$$

The discriminant $\Delta_{2}=4 k^{4}+4 k^{2}\left(1-k^{2}\right)\left(k^{2}+1\right)>0 \Rightarrow k^{2}<\frac{1+\sqrt{5}}{2}$.
By Vieta's formulas, we have $\left\{\begin{array}{l}x_{A}+x_{B}=\frac{2}{1-k^{2}} \\ x_{A} \cdot x_{B}=\frac{-\left(k^{2}+1\right)}{k^{2}\left(1-k^{2}\right)}\end{array}\right.$. Since $x_{A} \cdot x_{B}<0$, we have $k^{2}<1$, thus $k \in(0,1)$. According to the area formula of a triangle, we have:
$$
\begin{array}{l}
\left.S_{\triangle A O B}=\frac{1}{2}|| \begin{array}{ccc}
x_{A} & y_{A} & 1 \\
x_{B} & y_{B} & 1 \\
0 & 0 & 1
\end{array} \right\rvert\,=\frac{\left|x_{A} y_{B}-x_{B} y_{A}\right|}{2}=\frac{\left|x_{A}\left(k x_{B}+\frac{1}{k}\right)-x_{B}\left(k x_{A}+\frac{1}{k}\right)\right|}{2} \\
=\frac{\sqrt{\left(x_{A}+x_{B}\right)^{2}-4 x_{A} x_{B}}}{2 k}=\frac{\sqrt{\left(\frac{2}{1-k^{2}}\right)^{2} \frac{-4\left(k^{2}+1\right)}{k^{2}\left(1-k^{2}\right)}}}{2 k}=\frac{\sqrt{1+k^{2}-k^{4}}}{k^{2}\left(1-k^{2}\right)} \\
\end{array}
$$

Let $\mathrm{k}=\sin \alpha, \alpha \in(0, \pi)$, then
$$
\begin{array}{l}
S_{\triangle A O B}=\frac{\sqrt{1+\sin ^{2} \alpha-\sin ^{4} \alpha}}{\sin ^{2} \alpha \cos ^{2} \alpha}=\frac{\sqrt{1+\sin ^{2} \alpha \cos ^{2} \alpha}}{\sin ^{2} \alpha \cos ^{2} \alpha}=\sqrt{\frac{1}{(\sin \alpha \cos \alpha)^{4}}+\frac{1}{(\sin \alpha \cos \alpha)^{2}}} \\
=\sqrt{\frac{2^{4}}{\sin ^{4} 2 \alpha}+\frac{2^{2}}{\sin ^{2} 2 \alpha}} \geq \sqrt{2^{4}+2^{2}}=2 \sqrt{5}
\end{array}
$$

When $\alpha=\frac{\pi}{4}$, i.e., $\mathrm{k}=\frac{\sqrt{2}}{2}$, the equality holds.