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 √x2+y2=√x2+x4−10x2+25. In general minimization problems like this require calculus, but one elementary optimization method that sometimes works is completing the square. We have√x2+x4−10x2+25=√(x2−9/2)2+(25−81/4).
This expression is minimized when the square equals 0, i.e. when x=±3/√2. For this value of x, the distance is √25−814=√192. Hence the desired answer is 19+2=21.
