Answer: D
\((a^2 + b^2)(x^2 + y^2) = (ax + by)^2\) i.e., \(a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2\) = \(a^2x^2 + 2abxy + b^2y^2\) i.e., \(a^2y^2 + b^2x^2 = 2abxy\) i.e., \(a^2y^2 + b^2x^2 - 2abxy = 0\) i.e., \((ay)^2 - 2ay \cdot bx + (bx)^2 = 0\) i.e., \((ay - bx)^2 = 0\) i.e., \(ay - bx = 0\) i.e., \(bx = ay\) i.e., \(\frac{x}{y} = \frac{a}{b}\) \(\therefore x : y = a : b\)
\((a^2 + b^2)(x^2 + y^2) = (ax + by)^2\) i.e., \(a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2\) = \(a^2x^2 + 2abxy + b^2y^2\) i.e., \(a^2y^2 + b^2x^2 = 2abxy\) i.e., \(a^2y^2 + b^2x^2 - 2abxy = 0\) i.e., \((ay)^2 - 2ay \cdot bx + (bx)^2 = 0\) i.e., \((ay - bx)^2 = 0\) i.e., \(ay - bx = 0\) i.e., \(bx = ay\) i.e., \(\frac{x}{y} = \frac{a}{b}\) \(\therefore x : y = a : b\)