+184 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 on the graph is of the form (x,21x2−9), so the distance between the origin and this point is x2+(21x2−9)2. Squaring both sides, we get \begin{align*} a^2 &= x^2 + \left(\frac{1}{2}x^2-9\right)^2 \ &= x^2 + \frac{1}{4}x^4 - 9x^2 + 81 \ &= \frac{1}{4}x^4 - 8x^2 + 81. \end{align*}We can factor this as \begin{align*} a^2 &= \frac{1}{4}(x^2 - 8)^2 + \frac{324}{4} \ &= \frac{1}{4}(x^2 - 8)^2 + 81. \end{align*}The minimum value of 41(x2−8)2 is 0, which occurs when x2=8.
Therefore, the minimum value of a^2 is 81, so a^2=81.
