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# Please help!

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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

### 2+0 Answers

#1
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<<<< 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
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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