Graph drawing—Finding the points on a curve that are nearest to the origin

In summary, the problem involves finding the minimum distance to the origin on a tilted ellipse. This can be approached by using the Lagrange multiplier method, rotating to a new coordinate system, or using the AM-GM inequality. The solution can also be found by writing the equation of the curve in polar coordinates and applying domain restrictions to find the minimum distance.
  • #141
I thank you all for your patience:

Dick
Ray
WWGD
ehild
Phyton
epenguin
Nidum
Chestermiller
Mark44
tnich

It was a long ride
 
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  • #142
Karol said:
It was a long ride

Though a short distance. :oldsmile:
You took some long routes around where you saw some hopefully memorable scenery, like rotating axes, reciprocal equations amongst others.
Every method to find this shortest distance was trying to tell you at the same time the greatest distance, but this was always ignored. In order to fix the method of quadratic inequalities (#45) in your mind so that you might draw on it if you meet a suitable problem, I urged you find the largest distance using it. But as this hasn't been done I can't hold back from it .

Re-write the equation of your curve ##5x^2-6xy+5y^2=4## as
$$~~~~~~~~~~~~ 2(x^2+y^2) =4-3(x-y)^2$$
Then maximum of r^2 is 2 and the maximum of r is √2, which occurs when the square on the right is zero: there ##x=y=±1##.
 
  • #143
epenguin said:
Then maximum of r^2 is 2
Why?
 
  • #144
Karol said:
Why?

This has been explained to you many, many times already; just look at #4, #12, #38.
 
  • #145
I didn't know you can mix methods. i wouldn't have gotten to that
 
  • #146
Karol said:
Why?

Because any square must be ≥ 0, now can the RHS of the expression in #142 be greater than 4? Similar principle was involved in #45.
 
  • #147
Karol said:
I didn't know you can mix methods. i wouldn't have gotten to that

What dicatator told you that you cannot mix methods? Most good problem solvers mix methods all the time; learning to do that effectively should be part of any math education.
 
  • #148
It was a complex problem for me and i still feel unsure.
I thank you all again
This site and your attitude are invaluable for me
 
  • #149
It was far more satisfactory to me to interact with a student who saw the thing through to the end, told us all his results and reasoning including misunderstanding, than one who does not do these things. Or even who tells you 'thanks I have solved it' but you wonder if he has as well as he thinks.
 

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