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 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 (x-y) 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)/(1-k)]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? Thanks from greg1313
December 1st, 2017, 06:11 PM   #5
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Joined: Nov 2012

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Sure here is the question. Question 53
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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). Thanks from justusphung
 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 = 2|x|, where y isn't proportional to x and x - y isn't proportional to x + y.

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