10. 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 $A, B$ on the left and right branches, respectively. Find the minimum value of the area of $\triangle O A B$.
Solution
As shown in the figure, by symmetry, let point $P\left(t^{2}, 2 t\right)$ be in the first quadrant, then $l: t y=x+t^{2}$. Solving the system $\left\{\begin{array}{l}x^{2}-y^{2}=1, \\ t y=x+t^{2}\end{array}\right.$ $\Rightarrow\left(t^{2}-1\right) x^{2}-2 t^{2} x-t^{4}-t^{2}=0$.
Here $\Delta=4 t^{2}\left(t^{4}+t^{2}-1\right)>0$, let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, then combining $x_{1} x_{2}1$. Noting that the intersection point of line $l$ with the $y$-axis is $(0, t)$, we have $S_{\triangle O A B}=\frac{t}{2}\left|x_{1}-x_{2}\right|=\frac{t}{2} \cdot \frac{\sqrt{4 t^{2}\left(t^{4}+t^{2}-1\right)}}{t^{2}-1}=\frac{t^{2} \sqrt{t^{4}+t^{2}-1}}{t^{2}-1}$ $\xlongequal{m=t^{2}-1} \frac{(m+1) \sqrt{(m+1)^{2}+m}}{m}=\sqrt{\frac{\left(m^{2}+2 m+1\right)\left(m^{2}+3 m+1\right)}{m^{2}}}$ $=\sqrt{\left(m+\frac{1}{m}+2\right)\left(m+\frac{1}{m}+3\right)} \geqslant 2 \sqrt{5}$, equality holds when $m=1 \Rightarrow t=\sqrt{2}$. In summary, when $t= \pm \sqrt{2}$, the minimum area of $\triangle O A B$ is $2 \sqrt{5}$.