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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 \(a^2\).

 

 

Thank you so much for the help! I appreciate it!  

 Apr 6, 2021
 #1
avatar+602 
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The point is something like $(x,\frac{1}{2}x^2-9)$ for some constant value of $x$. The distance between this and the origin is $a=\sqrt{x^2+(\frac{1}{2}x^2-9)^2}$, and so $a^2=x^2+(\frac{1}{2}x^2-9)^2=x^2+\frac{1}{4}x^4-9x^2+81=\frac{1}{4}x^4-8x^2+81=\frac14(x^4-32x^2+324)$ and factor the inside as $(x^2+a)^2+b$ which has minimum value of $b$. Then multiply by $\frac14$ and finish. :)

 Apr 6, 2021
 #2
avatar+39 
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So 17 then?  

 Apr 6, 2021
 #3
avatar+602 
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Yep, believe so :)

 #4
avatar+39 
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It was correct! Thank you so much! The explanation cleared it up. 

 Apr 6, 2021

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