Example 2 (Question 14 of the 2008 National College Entrance Examination, Liberal Arts, First Paper) Given that the focus of the parabola $y=a x^{2}-1$ is the origin, then the area of the triangle formed by the three intersection points of the parabola with the two coordinate axes is $\qquad$ .
Solution: Since the parabola equation can be transformed into $x^{2}=\frac{1}{a}(y+$ $1)$, and it is known that its graph intersects the $y$-axis at point $C(0,-1)$, and the focus is the origin, by the geometric properties of the parabola, we know $\frac{p}{2}$ $=1 \Rightarrow p=2$, so $\frac{1}{a}=2 p=4 \Rightarrow a=\frac{1}{4}$. Therefore, the parabola equation is $y=\frac{1}{4} x^{2}-1$. The discriminant $\Delta=0-4 \cdot \frac{1}{4}(-1)=1$, hence by Theorem (2), $S=\frac{\Delta \sqrt{\Delta}}{8 a^{2}}=2$ is the required result.