# direct variation

• May 15th 2010, 01:29 AM
cakeboby
direct variation
given
$\displaystyle x^2 - y^2 \varpropto x^2 + y ^2$
prove that
(a)
$\displaystyle y\varpropto x$
(b)
$\displaystyle x-y \varpropto x+ y$
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(a)
$\displaystyle x^2 - y^2 = k (x^2 + y^2)$
$\displaystyle (1-k)x^2 = (k+1)y^2$
$\displaystyle x = \sqrt{\frac{k+1}{1-k}}y$
$\displaystyle y\varpropto x$

(b)
$\displaystyle x^2 - y^2 = k (x^2 + y^2)$
$\displaystyle (x-y)(x+y) = k ((x+y)^2 - 2xy)$
$\displaystyle x-y =k(x+y - \frac {2xy}{x+y})$
???????????
??????????????????

• May 15th 2010, 02:31 AM
sa-ri-ga-ma
http://www.mathhelpforum.com/math-he...2676b673-1.gif

Add y on both side and subtract y on both side and find (x+y) and (x-y) and take the ratio of (x+y)/(x-y).
• May 15th 2010, 09:33 PM
cakeboby
$\displaystyle x = \sqrt {\frac {k+1}{1-k}}y$
$\displaystyle x+y = \sqrt {\frac {k+1}{1-k}}y+y$
$\displaystyle x+y =( \sqrt {\frac {k+1}{1-k}}+1)y$

Cannot get it
• May 17th 2010, 12:29 AM
cakeboby
$\displaystyle x+y = \sqrt\frac{1+k}{1-k}y + y$
$\displaystyle x-y = \sqrt\frac{1+k}{1-k}y - y$
$\displaystyle \frac{(x+y)}{(x-y)} = \frac{\sqrt\frac{1+k}{1-k} + 1 }{\sqrt\frac{1+k}{1-k} - 1}$