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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!

 Feb 9, 2019
 #1
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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

 

 

cool  cool cool

 Feb 9, 2019

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