+0  
 
+2
342
1
avatar+598 

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 $a+b$.

michaelcai  Aug 11, 2017
 #1
avatar+90088 
+1

 

Let  u be the x coordinate of the point we seek....so u^2  - 5 is the y coordinate

 

And we are  seeking to minimize tthe distance, D, expressed as

 

D  = [ (u - 0)^2 + ( u^2  - 5 -  0 )^2 ]^(1/2)

 

D  = [ u^2  + u^4 - 10u^2  + 25 ]^(1/2)

 

D  = [ u^4  - 9u^2 + 25 ]^(1/2)

 

Take the derivative of this  and set it to 0

 

 

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

 

This will = 0 whenever [ 4u^3 - 18u]  = 0

 

Factor

 

u [ 4u^2  - 18]  = 0 

 

Setting both factors to 0

 

Either  u = 0  and  u^2 - 5  = 5    ....so one possible point is  (0, -5)  and this is 5 units from the origin

 

Or

 

4u^2  - 18  = 0

 

4u^2  = 18

 

u^2  = 18/4

 

u  = ± √ [18 /4]  = ± 3√2 / 2      and  u^2  - 5  =  18/4 - 5 =  18/4 - 20 / 4  = - 2/4 =   (-1/2)

 

So....the other points are  ( 3√2 / 4 , - 1/2 )   and  ( - 3√2 / 4 , - 1/2 ) 

 

And the distance that either one of these points is from the origin is given by

 

D  = √  [  (18/4)^2 - 9 (18/4)  + 25 ]  =  √  [ 324 - 648 + 400] / 4  =  √76 / 4  = 2√19/ 4  =  √19/2 ≈ 2.179

 

And this is the minimum distance from the origin to any point on the parabola

 

So...... √a / b  =    √19 / 2      and   a +  b  =  19  + 2  = 21

 

 

 

cool cool cool

CPhill  Aug 11, 2017
edited by CPhill  Aug 12, 2017

20 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.