The smallest distance between the origin and a point on the graph of y=(1/sqrt(2))((x^2)-3) 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 .

Thanks!

Guest Feb 9, 2019

#1**+1 **

Let the point on the graph be [ a, (1/√2)(a^2 - 3) ]

So....if D is minimized....then so is D^2

So we have that

D^2 = (a - 0)^2 + [ (1/√2)(a^2 - 3) - 0)]^2

D^2 = a^2 + (1/2)(a^2- 3)^2 take the derivative of this and set to 0

D^2 ' = 2a + (a^2 - 3)(2a) = 0

D^2 ' = 2a [ 1 + (a^2 - 3 ] = 0

This implies that either

2a = 0

a = 0

Or that

1 + a^2 - 3 = 0

a^2 - 2 = 0

a = ±√2

When a = 0, D^2 = 0^2 + (1/2)(0^2- 3)^2 = 9/2

When a = -√2 , D^2 = (-√2)^2 + (1/2) [ (-√2)^2 - 3 ]^2 = 5/2

The same result will resullt when a = √2

So D will be √[5/2] = √2 * √5 / 2 = √10 / 2

And a + b = 12

CPhill Feb 9, 2019