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December 1st, 2017, 05:42 AM   #1
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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)?
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December 1st, 2017, 06:00 AM   #2
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Perhaps "(x² + y²)" should have been "(x + y)²".
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December 1st, 2017, 06:22 AM   #3
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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.
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December 1st, 2017, 02:08 PM   #4
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Can you provide the question in full, in the exact form in which it was originally given?
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December 1st, 2017, 06:11 PM   #5
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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.
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December 1st, 2017, 10:28 PM   #6
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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).
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December 2nd, 2017, 01:57 AM   #7
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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.
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December 2nd, 2017, 02:33 AM   #8
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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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