11. Given the parabola $y=a x^{2}$, lines $l_{1}, l_{2}$ both pass through the point $(1,-2)$ and are perpendicular to each other. If the parabola intersects with at least one of the lines $l_{1}$, $l_{2}$, find the range of the real number $a$.
11. Solution: If lines $l_{1}, l_{2}$ are parallel to the coordinate axes, then the parabola intersects with at least one of the lines $l_{1}, l_{2}$. Let $l_{1}: y+2=k(x-1), l_{2}: y+2=-\frac{1}{k}(x-1)$. Without loss of generality, assume $k>0$. By eliminating $y$ from $y+2=k(x-1)$ and $y=a x^{2}$, we get $a x^{2}-k x+(k+2)=0$, with the discriminant $\Delta_{1}=k^{2}-4 a k-8 a$.
Substituting $-\frac{1}{k}$ for $k$ in the above equation, we get the discriminant $\Delta_{2}=\left(-\frac{1}{k}\right)^{2}-4 a\left(-\frac{1}{k}\right)-8 a$. According to the problem, $\Delta_{1} \geqslant 0$ or $\Delta_{2} \geqslant 0$ must hold.
From $\Delta_{1} \geqslant 0$, we get $k^{2}-4 a k-8 a \geqslant 0$.
From $\Delta_{2} \geqslant 0$, we get $8 a k^{2}-4 a k-1 \leqslant 0$.
If $a<0$, then (1) must hold, now assume $a>0$,
Let $f(k)=k^{2}-4 a k-8 a, g(k)=8 a k^{2}-4 a k-1$,

Then $y=f(k), y=g(k)$ are both quadratic functions of $k$, and their graphs open upwards, with the axes of symmetry to the right of the $y$-axis. Note that $f(0)=-8 a<0$, so there exists $\alpha>0$ such that $f(\alpha)=0$,
i.e., $\alpha^{2}-4 a \alpha-8 a=0$.
When $k \geqslant \alpha$, $f(k) \geqslant 0$, i.e., (1) holds,
When $0<k<\alpha$, $f(k)<0$, at this time (1) does not hold,
Then (2) must hold, so $g(\alpha) \leqslant 0$ must be true,
Substituting in, we get $8 a \alpha^{2}-4 a \alpha-1 \leqslant 0$. (4)
From (3) and (4), we get $(8 a-1)\left(a^{2}+1\right) \leqslant 0$, solving this gives $0<a \leqslant \frac{1}{8}$,
In summary, the range of real number $a$ is $(-\infty, 0) \cup\left(0, \frac{1}{8}\right]$.