Expert: Paul Klarreich - 12/2/2008

Question
QUESTION: Find the point(s) on the hyperbola x²-y²=4 that is closest to the point (1,3).

ANSWER: Questioner:   Eric
Category:  Calculus
Private:  No
 
Subject:  Min/Max problem
Question:  Find the point(s) on the hyperbola x²-y²=4 that is closest to the point (1,3).
............................
Hi, Eric,

The trick here is to minimize D^2, so you don't have too many radicals to worry about.

Let (x,y) be the point.

D^2[from (x,y) to (1,3)] = (x - 1)^2 + (y - 3)^2

Constraint:  x^2  - y^2 = 4

You will have to deal with y = sqrt(x^2 - 4)


D^2 = x^2 - 2x + 1 + y^2     - 6  y          + 9

D^2 = x^2 - 2x + 1 + x^2 - 4 - 6 sqrt(x^2 - 4) + 9

D^2 = 2x^2 - 2x  + sqrt(x^2 - 4)

         x
D^2' = 4x - 2 + ------------
         sqrt(x^2-4)


Set that = 0:
         x
4x - 2 + ------------ = 0
         sqrt(x^2-4)


This is going to be messy, so I'll leave it to you.


---------- FOLLOW-UP ----------

QUESTION:      x
D^2' = 4x - 2 + ------------
         sqrt(x^2-4)


Set that = 0:
         x
4x - 2 + ------------ = 0
        sqrt(x^2-4)


What do I do after I set that equal to zero


Answer
Questioner:  Eric
Private: no
Subject:   

 
Question:  
QUESTION: Find the point(s) on the hyperbola x²-y²=4 that is closest to the point (1,3).

ANSWER: Questioner:   Eric
Category:  Calculus
Private:  No

Subject:  Min/Max problem
Question:  Find the point(s) on the hyperbola x²-y²=4 that is closest to the point (1,3).
............................
Hi, Eric,

The trick here is to minimize D^2, so you don't have too many radicals to worry about.

Let (x,y) be the point.

D^2[from (x,y) to (1,3)] = (x - 1)^2 + (y - 3)^2

Constraint:  x^2  - y^2 = 4

You will have to deal with y = sqrt(x^2 - 4)


D^2 = x^2 - 2x + 1 + y^2     - 6  y          + 9

D^2 = x^2 - 2x + 1 + x^2 - 4 - 6 sqrt(x^2 - 4) + 9

D^2 = 2x^2 - 2x  + sqrt(x^2 - 4)

         x
D^2' = 4x - 2 + ------------
         sqrt(x^2-4)


Set that = 0:
         x
4x - 2 + ------------ = 0
        sqrt(x^2-4)


This is going to be messy, so I'll leave it to you.


---------- FOLLOW-UP ----------

QUESTION:      x
D^2' = 4x - 2 + ------------
         sqrt(x^2-4)


Set that = 0:
         x
4x - 2 + ------------ = 0
       sqrt(x^2-4)


What do I do after I set that equal to zero


   x
------------ = 2 - 4x
sqrt(x^2-4)


  x  
-------- = sqrt(x^2 - 4)
(2 - 4x)

  x^2
----------- = x^2 - 4
(2 - 4x)^2


As I said, this is going to be messy, so I'll leave it to you.