Question 176: Let point $\mathrm{C}$ be a moving point on the parabola $\mathrm{y}^{2}=2 \mathrm{x}$. Draw two tangents $C A$ and $C B$ from $\mathrm{C}$ to the circle $(\mathrm{x}-1)^{2}+\mathrm{y}^{2}=1$, intersecting the negative half-axis of the $y$-axis at $A$ and the positive half-axis of the $y$-axis at $B$. Find the minimum value of the area of $\triangle A B C$.
Question 176, Solution: Let point $\mathrm{C}\left(2 \mathrm{a}^{2}, 2 \mathrm{a}\right)$, then $\mathrm{a}>1$. Suppose the equation of the tangent line is $\mathrm{y}=\mathrm{kx}+\mathrm{b} \hat{5}$, then $2 \mathrm{a}=$ $2 a^{2} \cdot k+b$, solving for $b$ gives $b=2 a(1-a k)$, so the equation of the tangent line is:
$$
\mathrm{y}=\mathrm{kx}+2 \mathrm{a}(1-\mathrm{ak})
$$

Solving the system of equations: $\left\{\begin{array}{c}y=k x+2 a(1-a k) \\ (x-1)^{2}+y^{2}=1\end{array}\right.$, we get:
$$
\left(k^{2}+1\right) x^{2}+2[2 \mathrm{ak}(1-\mathrm{ak})-1] \mathrm{x}+4 \mathrm{a}^{2}(1-\mathrm{ak})^{2}=0
$$

Since it is a tangent line, its discriminant $\Delta=4[2 \mathrm{ak}(1-\mathrm{ak})-1]^{2}-4\left(\mathrm{k}^{2}-1\right)\left[4 \mathrm{a}^{2}(\mathrm{a}-\mathrm{ak})^{2}\right]=0$, simplifying we get:
$$
4 a^{2}\left(a^{2}-1\right) k^{2}+4 a\left(1-2 a^{2}\right) k+4 a^{2}-1=0
$$

Thus, $\left|k_{1}-k_{2}\right|=\sqrt{\left(k_{1}+k_{2}\right)^{2}-4 k_{1} k_{2}}=\sqrt{\left[\frac{2 a^{2}-1}{a\left(a^{2}-1\right)}\right]^{2}-\frac{4 a^{2}-1}{a^{2}\left(a^{2}-1\right)}}=\frac{1}{a^{2}-1}$, so $A B=$ $2 \mathrm{a}^{2}\left|\mathrm{k}_{1}-\mathrm{k}_{2}\right|=\frac{2 \mathrm{a}^{2}}{\mathrm{a}^{2}-1}$, thus the area of $\triangle A B C$ is:
$$
S_{\triangle A B C}=\frac{\left(2 a^{2}\right)\left(\frac{2 a^{2}}{2}\right)}{2^{2}-1}=\frac{2 a^{4}}{a^{2}-1}
$$

Let $t=a^{2} y-1$ then $t>0$, and $S_{\triangle A B C}=\frac{2(t+1)^{2}}{t}=8+\frac{2(t-1)^{2}}{t} \geq 8$, so the minimum value of $S_{\triangle A B C}$ is 8.