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 $a^2$.

The point is something like $(x,\frac{1}{2}x^2-9)$ for some constant value of $x$. The distance between this and the origin is $a=\sqrt{x^2+(\frac{1}{2}x^2-9)^2}$, and so $a^2=x^2+(\frac{1}{2}x^2-9)^2=x^2+\frac{1}{4}x^4-9x^2+81=\frac{1}{4}x^4-8x^2+81=\frac14(x^4-32x^2+324)$ and factor the inside as $(x^2+a)^2+b$ which has minimum value of $b$. Then multiply by $\frac14$ and finish. :)

So 17 then?  

It was correct! Thank you so much! The explanation cleared it up.