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 D=√x2+y2.
So, if
y=x2−5, thenD=√x2+(x2−5)2=√x4−9x2+25.
Now complete the square, (add and subtract (9/2)^2),
D=√x4−9x2+(81/4)+(100/4)−(81/4)=√{x2−(9/2)}2+(19/4).
This will take its smallest value
√19/4=√19/2, when x2=9/2.
Tiggsy
