The smallest distance between the origin and a point on the graph of \(y=\frac{1}{\sqrt{2}}\left(x^2-3\right)\) can be expressed as \(\sqrt{a}/b\), where \(a\) and \(b\) are positive integers such that \(a\) is not divisible by the square of any integer greater than one. Find \(a+b\).
+129850
Let the point be ( x , (x^2 - 3) / sqrt(2) )
D = sqrt ( x^2 + (x^2 - 3)^2 /2 ) take the derivative and set to 0
D' = (2x + 2x(x^2 - 3) ) / sqrt ( x^2 + (x^2 - 3)^2 /2 ) = 0
2x + 2x^3 - 6x = 0
2x^3 - 4x = 0
2x ( x^2 - 2) = 0
x = 0, x = sqrt (2) or x = -sqrt (2)
When x = 0
D = sqrt (0 + 9/2) = sqrt (9/2)
When x = ±sqrt (2)
D = sqrt ( 2 + 1/2) = sqrt (5/2) = sqrt (10)/2
a = 10 and b = 2
So
a + b = 12
