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 $a+b$.
The distance from a point (x, y) to the origin is given by \(\displaystyle D = \sqrt{x^{2}+y^{2}}.\)
So, if
\(\displaystyle y=x^{2}-5,\text{ then}\\D=\sqrt{x^{2}+(x^{2}-5)^{2}}=\sqrt{x^{4}-9x^{2}+25}.\)
Now complete the square, (add and subtract (9/2)^2),
\(\displaystyle D = \sqrt{x^{4}-9x^{2}+(81/4)+(100/4)-(81/4)}\\=\sqrt{\{x^{2}-(9/2)\}^{2}+(19/4)}.\)
This will take its smallest value
\(\displaystyle \sqrt{19/4}=\sqrt{19}/2, \text{ when }x^{2}=9/2.\)
Tiggsy
