+0  
 
0
60
1
avatar

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 .

Guest Apr 19, 2018
 #1
avatar+87294 
+1

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

CPhill  Apr 19, 2018

12 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.