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The smallest distance between the origin and a point on the parabola y = x^2 -5 can be expressed as "sqrt(a)/b," where "a" is not divisible by the square of any prime. Find a+b .

 Apr 19, 2018
 #1
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y  = x^2  - 5

 

Let the point on the parabola  be  (x, x^2 - 5)

 

The distance, D, between this point and the origin can be represented by :

 

D  = sqrt  [  x^2  +  ( x^2 - 5)^2 ]

D  = sqrt [ x^2 + x^4 - 10x^2 + 25]

D  = [ x^4 - 9x^2 + 25 ]^(1/2)

 

Take the derivative of this and set to 0

 

D'  =  [(1/2] [(x^4 - 9x^2 + 25 ]^(-1/2)* (4x^3 - 18x)  = 0

 

This will equal 0  when  (4x^3 - 18x)  = 0       factor  this

 

2x (2x^2 - 9)  = 0

Set both factors to 0   and solve for x

 

2x  =  0                     2x^2  - 9  = 0

x  = 0                         2x^2  =  9

reject                           x^2 = 9/2

                                   

 

So....the distance, D, is

 

D  = sqrt  [ (9/2)^2 - 9(9/2) + 25 ]  =  sqrt [  81/4 - 81/2 + 25 ] =

 

sqrt [ 81/ 4 - 162/4  + 100/4]  =

 

sqrt  [ 81 - 162 + 100]  / 2  =

 

sqrt  [ 19 ] /2

 

So....a +  b  =   19 + 2   =  21

 

 

cool cool cool

 Apr 19, 2018

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