15. Given the parabola $y=a x^{2}$ passes through the point $P(-1,1)$, a line $l$ with a positive slope is drawn through the point $Q\left(-\frac{1}{2}, 0\right)$ intersecting the parabola at points $M, N$ (point $M$ is between $Q$ and $N$). A line parallel to the $x$-axis is drawn through point $M$, intersecting $O P$ at $A$ and $O N$ at $B$. The areas of $\triangle P M A$ and $\triangle O A B$ are denoted as $S_{1}$ and $S_{2}$, respectively. Compare $S_{1}$ with $3 S_{2}$, and explain the reasoning.
Parse the parabola $y=a x^{2}$ passing through point $P(-1,1)$, we get $a=1$, so the equation of the parabola is $y=x^{2}$. Let the equation of line $l$ be $y=k\left(x+\frac{1}{2}\right)$ (where $k>0$), from $\left\{\begin{array}{l}y=k\left(x+\frac{1}{2}\right) \\ y=x^{2}\end{array}\right.$, we get $2 x^{2}-2 k x-k=0$. Let $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then $x_{1}1-\frac{1}{4}=\frac{3}{4}$. Also, since $S_{1}=\frac{1}{2}|A M| \cdot\left(1-y_{1}\right), S_{2}=\frac{1}{2}|A B| \cdot y_{1}$, we have $S_{1}>3 S_{2}$.