Question-31
If two distinct non-zero real variables \(x\) and \(y\) are such that \((x+y)\) is proportional to \((x-y)\) then the value of \(\cfrac{x^2+y^2}{x^2-y^2}\) is
Let \(\displaystyle x+y=k( x-y)\), with \(\displaystyle x,y\neq 0\) and \(\displaystyle x,y\in \mathbb{R}\), then:
\[ \begin{aligned} x+y & =k( x-y)\\ & \\ ( 1-k) x & =-( 1+k) y\\ & \\ \frac{x}{y} & =\frac{k+1}{k-1} \end{aligned} \]
provided \(\displaystyle k\neq 1\). \(\displaystyle k\) can’t be equal to \(\displaystyle 1\) since that would lead to \(\displaystyle y=0\), which is not possible as per the constraints in the question. We now see that \(\displaystyle x\) is proportional to \(\displaystyle y\) with some other proportionality constant. Now:
\[ \begin{aligned} \frac{x^{2} +y^{2}}{x^{2} -y^{2}} & =\frac{\frac{x^{2}}{y^{2}} +1}{\frac{x^{2}}{y^{2}} -1}\\ & \\ & =\frac{k^{2} +1}{2k} \end{aligned} \]
We see that \(\displaystyle \frac{x^{2} +y^{2}}{x^{2} -y^{2}}\) is a constant.