Question: What is the smallest distance between the origin and a point on the graph of y=1√2(x2−18) ?
What I tried: I want to minimize √x2+(1√2(x2−18))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,√17,−√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 √35/2. This was wrong. Can someone help? Thanks!
+1382
<<<< 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
x2+{1√2(x2−18)}2
=12(2x2+x4−36x2+324)=12(x4−34x2+324)=12(x4−34x2+289+324−289)=12{(x2−17)2+35}.
That has a minimum value of 35/2 occurring when x squared is equal to 17.
So the minimum distance is√35/2 .
