যদি \(\left(\cfrac{1}{x}-\cfrac{1}{y}\right)∝\cfrac{1}{x-y}\) হয় তবে দেখাও যে, \((x^2+y^2)∝xy\) ।
Madhyamik 2019
\(\left(\cfrac{1}{x}-\cfrac{1}{y}\right)∝\cfrac{1}{x-y}\)
বা, \(\left(\cfrac{1}{x}-\cfrac{1}{y}\right)=k.\cfrac{1}{x-y}\) [\(k\) অশূন্য ভেদ ধ্রুবক]
বা, \(\cfrac{y-x}{xy}=k.\cfrac{1}{x-y}\)
বা, \(\cfrac{-(x-y)}{xy}=k.\cfrac{1}{x-y}\)
বা, \(-(x-y)^2=kxy\)
বা, \((x-y)^2=-kxy\)
বা, \(x^2+y^2-2xy=-kxy\)
বা, \(x^2+y^2=2xy-kxy\)
বা, \(x^2+y^2=(2-k)xy\)
বা, \(\cfrac{x^2+y^2}{xy}=(2-k)=\) ধ্রুবক
\(\therefore (x^2+y^2)∝xy\) (প্রমাণিত)