+618 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 $a+b$.
+128474
Let u be the x coordinate of the point we seek....so u^2 - 5 is the y coordinate
And we are seeking to minimize tthe distance, D, expressed as
D = [ (u - 0)^2 + ( u^2 - 5 - 0 )^2 ]^(1/2)
D = [ u^2 + u^4 - 10u^2 + 25 ]^(1/2)
D = [ u^4 - 9u^2 + 25 ]^(1/2)
Take the derivative of this and set it to 0
D' = (1/2) [ u^4 - 9u^2 + 25 ]^(-1/2) * [ 4u^3 - 18u] = 0
This will = 0 whenever [ 4u^3 - 18u] = 0
Factor
u [ 4u^2 - 18] = 0
Setting both factors to 0
Either u = 0 and u^2 - 5 = 5 ....so one possible point is (0, -5) and this is 5 units from the origin
Or
4u^2 - 18 = 0
4u^2 = 18
u^2 = 18/4
u = ± √ [18 /4] = ± 3√2 / 2 and u^2 - 5 = 18/4 - 5 = 18/4 - 20 / 4 = - 2/4 = (-1/2)
So....the other points are ( 3√2 / 4 , - 1/2 ) and ( - 3√2 / 4 , - 1/2 )
And the distance that either one of these points is from the origin is given by
D = √ [ (18/4)^2 - 9 (18/4) + 25 ] = √ [ 324 - 648 + 400] / 4 = √76 / 4 = 2√19/ 4 = √19/2 ≈ 2.179
And this is the minimum distance from the origin to any point on the parabola
So...... √a / b = √19 / 2 and a + b = 19 + 2 = 21
