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# need help

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What is the smallest distance between the origin and a point on the graph of  $$y=\dfrac{1}{\sqrt{2}}\left(x^2-8\right)$$?

Little confused. How to proceed? I was thinking find (x,y) formula for the equation, then use distance formula?

Apr 14, 2022

#1
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This question has already been posted: https://web2.0calc.com/questions/1-what-is-the-smallest-distance-between-the-origin

Apr 14, 2022
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(deleted)

Apr 14, 2022
edited by Vinculum  Apr 14, 2022
#3
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Hi, I would not like answers, just how to proceed.

Guest Apr 14, 2022
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Let  the point we seek  =    (x , y)  =   (x ,  (x^2 -8) /sqrt 2 )

If we minimize  the distance, d,   then we also minimize the distance  d^2

So....using the square of the distance  formula   between the origin  and this point   we have  that

d^2  =      ( x - 0)^2   +  ( (x^2 -8) /sqrt 2 - 0 )^2      simplify

d^2   =  x^2  +   (1/2)  (  x^4 - 16x^2 + 64 )

d^2  =  x^2  + (1/2)x^4 - 8x^2 + 32

d^2  =  (1/2)x^4 -7x^2 + 32          take the derivative of this  function  and  set to 0

2x^3 - 14x   =   0        factor

2x ( x^2 - 7)   = 0

The second factor set to 0  is  what we want

x^2  - 7  = 0

x^2  = 7

x =sqrt (7)

And    y =  [ (sqrt (7) )^2  - 8 ] / sqrt 2  =  -1 /sqrt 2

So the smallest distance  =    sqrt  [ (sqrt 7)^2  +  (-1 /sqrt 2)^2 ]  =    sqrt  [ 7 + 1/2]  = sqrt (7.5)

Here's a graph : https://www.desmos.com/calculator/49brfcvcvv

Note that, because we have a parabola that is symmetric to the origin, (-sqrt (7) , -1/sqrt 2)  also produces a point on the parabola that has the same minimum  distance  from the origin   Apr 14, 2022
edited by CPhill  Apr 14, 2022