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

 May 4, 2021
 #1
avatar+32410 
+3

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

 

 

(where I have distance from origin, I should have put distance squared from origin!)

 May 4, 2021
edited by Alan  May 4, 2021
 #2
avatar+2135 
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Origin is: (0, 0)

Point in parabola: (x, y)

Distance: sqrt(x^2 + y^2)

 

sqrt(x^2 + y^2)

sqrt(x^2 + (x^2 - 2)^2)

sqrt(x^2 + x^4 - 4x^2 + 4)

sqrt(x^4 - 3x^2 + 4)

Now, we have to find the smallest positive value of x^4 - 3x^2 + 4. 

Not quite sure how to do that. 

 

=^._.^=

 May 4, 2021
 #3
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+1

Can use calculas as in #1 above, or altermatively completing the square works.

 

\(\displaystyle x^{4}-3x^{2}+4 = x^{4}-3x^{2}+(9/4) + 4 -(9/4)\\ = (x^{2}-3/2)^{2}+7/4,\)

 

so the minimum distance is \(\displaystyle \sqrt{7/4}\) from either \(\displaystyle x=\pm\sqrt{3/2}.\)

 May 4, 2021
 #4
avatar+2135 
0

Ohhh, I can't believe I didn't think of completing the square. 

Nice job :))

 

=^._.^=

catmg  May 4, 2021

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