If \(\frac{x}{y} = \frac{b + 3}{b - 3}\), then find the value of \(\frac{x^2 + y^2}{x^2 - y^2}\).
\(\frac{b^2 + 9}{6b}\)
Since only the ratio x : y matters, take x = b + 3 and y = b – 3 (any common multiplier cancels in the required fraction).
Using \((a + c)^2 = a^2 + 2ac + c^2\): \(x^2 + y^2 = (b + 3)^2 + (b - 3)^2 = 2b^2 + 18 = 2(b^2 + 9)\).
Using \(a^2 - c^2 = (a + c)(a - c)\): \(x^2 - y^2 = (2b)(6) = 12b\).
Therefore \(\frac{x^2 + y^2}{x^2 - y^2} = \frac{2(b^2 + 9)}{12b} = \frac{b^2 + 9}{6b}\).
Check: take b = 5, so x : y = 8 : 2 = 4 : 1; then \(\frac{16 + 1}{16 - 1} = \frac{17}{15}\), and \(\frac{25 + 9}{30} = \frac{34}{30} = \frac{17}{15}\), the same value.
Hence, \(\frac{b^2 + 9}{6b}\).
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