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Question: What is the smallest distance between the origin and a point on the graph of \(y=\frac{1}{\sqrt2}(x^2-18)\) ?

 

What I tried: I want to minimize \(\sqrt{x^2+(\frac{1}{\sqrt2}(x^2-18))^2}\) . Since this must be nonnegative, I figured I only really needed to minimize what's inside the square root. I took the derivative and set it to 0 and got x values of \(0, \sqrt{17},-\sqrt{17}\). When I graphed y, it appeared that the positive and negative root 17 were the minimum x values. I plugged x in to find y, and used the distance formula to get \(\sqrt{35/2}\). This was wrong. Can someone help? Thanks!

 Jul 1, 2023
 #1
avatar+832 
+3

 

<<<< it appeared that the positive and negative root 17 were the minimum x values >>>>   

 

I approximated 1/sqrt(2) as 0.707 and used Desmos to graph y = 0.707(x2 – 18)  

My parabola crossed the x-axis near +4.2 ... that's close enough to sqrt(17) that    

I'd say we got the same answer there.  

 

I believe, however, that that is not the shortest distance to the origin.  

 

If you look at the graph, you see that the curve is slanting outward and 

crosses the x-axis at an angle.  The measurement to that intersection  

from the origin would be a line on top of the x-axis itself.  

 

The shortest distance from the origin to the curve has to be the line that's     

perpendicular to the curve.  That's going to be somewhere below the x-axis.    

Eyeballing the graph on Desmos, which is an approximation, the point seems    

to be somewhere close to (4, –2) on the positive side.  I don't know how to do   

the calculations that would determine it accurately.   

.

 Jul 2, 2023
 #2
avatar+397 
+3

I used a different method and arrived at the same result as you.

The square of the distance from the origin to the parabola is

\(\displaystyle x^{2}+\left\{\frac{1}{\sqrt{2}}(x^{2}-18)\right\}^{2}\)

\(\displaystyle =\frac{1}{2}(2x^{2}+x^{4}-36x^{2}+324) \\ \displaystyle =\frac{1}{2}(x^{4}-34x^{2}+324) \\ \displaystyle = \frac{1}{2}(x^{4}-34x^{2}+289+324-289) \\ \displaystyle = \frac{1}{2}\{(x^{2}-17)^{2}+35\}.\)

That has a minimum value of 35/2 occurring when x squared is equal to 17.

So the minimum distance is\(\displaystyle \sqrt{35/2}\)  .                                                                   

 Jul 2, 2023

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