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Find the smallest distance between the origin and a point on the parabola y = x^2 -2.

 Nov 17, 2020
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y = x^2  - 2

 

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

 

Using the distance formula  we  have  that

 

D =   [ x^2  +  ( x^2 - 2)^2  ]^(1/2)       simplify

 

D = [ x^2  + x^4  - 4x^2 + 4] ^(1/2)

 

D =  [ x^4 - 3x^2 + 4 ) ^(1/2)       take the  derivative of this  and set to 0

 

D'  = [ 4x^3 + 6x ] / [ x^4 - 3x^2 + 4 ] ^(1/2)  = 0

 

So

 

4x^3  + 6x   =  0

 

2x (2x^2 - 3)  = 0

 

2x  = 0                                   2x^2 - 3  = 0

x  = 0   reject                          2x^2  = 3

                                                 x^2 = 3/2

                                                 x  = ±sqrt (3/2)

 

When x =  sqrt (3/2)    then y  =  [sqrt (3/2)] ^2 - 2 =  -1/2

When x = -sqrt (3/2)  y has the same value = -1/2

 

So....the shortest distance in either case is

 

sqrt  ( 3/2  + 1/2)  = sqrt (2)

 

See the graph, here : https://www.desmos.com/calculator/i36wydpg70

 

 

cool cool cool

 Nov 17, 2020

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