December 1st, 2017, 05:42 AM  #1 
Member Joined: Nov 2012 Posts: 65 Thanks: 1  Variation
If (x^2  y^2) is directly proportional to (x^2 + y^2) Then how to show that (xy) is also directly varied as (x+y)? Thanks. 
December 1st, 2017, 06:00 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
Perhaps "(x² + y²)" should have been "(x + y)²".

December 1st, 2017, 06:22 AM  #3 
Member Joined: Nov 2012 Posts: 65 Thanks: 1 
If it is (x+y)^2, then the proof is quite straight forward. So, you are pretty sure that there is a mistake with the question itself? As the first part of the question is to show that y is directly proportional to x, I did it by showing that x = +/ sqrt[(1+k)/(1k)]y, but the second part seems impossible to prove.

December 1st, 2017, 02:08 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
Can you provide the question in full, in the exact form in which it was originally given?

December 1st, 2017, 06:11 PM  #5 
Member Joined: Nov 2012 Posts: 65 Thanks: 1 
Sure here is the question. Question 53
Last edited by justusphung; December 1st, 2017 at 06:27 PM. 
December 1st, 2017, 10:28 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
Let c = √((1  k)/(k + 1)). You showed that y = ±cx, which isn't the same as y = cx. If y = cx, x  y = ((1  c)/(1 + c))(x + y). 
December 2nd, 2017, 01:57 AM  #7 
Member Joined: Nov 2012 Posts: 65 Thanks: 1 
Thx for your prompt reply. Part b cannot be solved by using the information from the question, but can be done by using the answer in part a.

December 2nd, 2017, 02:33 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 19,870 Thanks: 1833 
Yes. If, for example, x²  y² = (3/5)(x² + y²), you get y² = 4x². You are okay (y is proportional to x) if y = 2x or if y = 2x, but there are other possibilities, such as y = 2x, where y isn't proportional to x and x  y isn't proportional to x + y. 

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