Question:
The smallest distance between the origin and a point on the graph of $y=\frac{1}{2}x^2-9$ can be expressed as $a$.  Find $a^2$.

Answer:
By the distance formula, we are trying to minimize $\sqrt{x^2+y^2}=\sqrt{x^2+\frac{1}{4}x^4-9x^2+81}$. In general, minimization problems like this require calculus, but one optimization method that sometimes works is to try to complete the square.  Pulling out a factor of $\frac{1}{4}$ from under the radical, we have \begin{align*}
\frac{1}{2}\sqrt{4x^2+x^4-36x^2+324}&=\frac{1}{2}\sqrt{(x^4-32x^2+256)+68} \\
&= \frac{1}{2}\sqrt{(x^2-16)^2+68}
\end{align*}This last expression is minimized when the square equals $0$, that is, when $x^2=16$. Then the distance is $\frac{\sqrt{68}}{2}=\sqrt{17}$.  Hence the desired answer is $\sqrt{17}^2 = \boxed{17}$.