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!
+128408 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
