The smallest distance between the origin and a point on the parabola $y=x^2-5$ can be expressed as $\sqrt{a}/b$, where $a$ is not divisible by the square of any prime. Find $a+b$.
+37 By the distance formula, we are trying to minimize \(\sqrt{x^2+y^2}=\sqrt{x^2+x^4-10x^2+25}\). In general minimization problems like this require calculus, but one elementary optimization method that sometimes works is completing the square. We have\(\sqrt{x^2+x^4-10x^2+25}=\sqrt{(x^2-9/2)^2+(25-81/4)}.\)
This expression is minimized when the square equals \(0\), i.e. when \(x=\pm 3/\sqrt{2}.\) For this value of \(x\), the distance is \(\sqrt{25-\frac{81}{4}}=\frac{\sqrt{19}}{2}.\) Hence the desired answer is \(19+2=\boxed{21}\).
