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The smallest distance between the origin and a point on the graph of y = $\frac{1}{2}x^2$ - 9 can be expressed as a. Find a^2.

 Jun 30, 2022
 #1
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\( y = $\frac{1}{2}x^2$ - 9 \)

 

Call the point we are looking  for (c , c^2/2 - 9 )

 

Using the square of the  distance  formula  we have

 

D^2  =  ( c - 0)^2  + ( c/^2 / 2  - 9  - 0  )^2

 

D^2  = c^2  + c^4/4 - 9c^2 + 81

 

D^2  =  c^4 /4 - 8c^2 +  81           take the derivative

 

D^2 '   =   c^3 -16c            set to 0

 

c^3 - 16c  = 0

 

c ( c^2 - 16)  = 0

 

The second factor  set  = 0   is what we want

 

c^2  = 16

 

Two vaues of c  will  minimize the distance c = 4 or c  =-4......choose  c = 4

 

And y  = 4^2 / 2 - 9  =    -1

 

The point that minimizes the distance is  ( 4 , -1) 

 

a^2  =  4^2 + (-1)^2  =    17

 

 

cool cool cool

 Jun 30, 2022
edited by CPhill  Jun 30, 2022
 #2
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+1

Thank you!!!!!!!!

Guest Jun 30, 2022

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