+0  
 
0
216
1
avatar

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 \(a+b\).

 Mar 16, 2019
 #1
avatar+111438 
0

 

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

 

cool cool cool

 Mar 16, 2019

7 Online Users