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!
+1237
<<<< 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.
.
+397 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\(\displaystyle \sqrt{35/2}\) .
