If $(a + b) : (a - b) = 1 : 5$ , find $(a^{2} - b^{2}) : (a^{2} + b^{2}) .$

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Solution

Given : $(a + b) : (a - b) = 1 : 5$
$⟹ \frac{a + b}{a - b} = \frac{1}{5}$
$⟹ 5 a + 5 b = a - b$
$4 a = - 6 b$
$a = - \frac{3}{2}$
Consider, $\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$
$= \frac{(\frac{- 3}{2})^{2} - b^{2}}{(\frac{- 3}{2})^{2} + b^{2}}$
$= \frac{(\frac{9}{4}) - b^{2}}{(\frac{9}{4}) + b^{2}}$
$= \frac{9 b^{2} - 4 b^{2}}{9 b^{2} + 4 b^{2}}$
$= \frac{5 b^{2}}{13 b^{2}}$
$= \frac{5}{13}$
Hence, $(a^{2} - b^{2}) : (a^{2} + b^{2}) = 5 : 13$

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