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Mathematics

The mean proportion between x−yx+y and x2y2x2−y2\dfrac{x - y}{x + y} \text{ and } \dfrac{x^2y^2}{x^2 - y^2} is :

  1. xyx+y\dfrac{xy}{x + y}

  2. xyx−y\dfrac{xy}{x - y}

  3. x−yxy\dfrac{x - y}{xy}

  4. x+yxy\dfrac{x + y}{xy}

Ratio Proportion

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Answer

Let a be the mean proportion between x−yx+y and x2y2x2−y2\dfrac{x - y}{x + y} \text{ and } \dfrac{x^2y^2}{x^2 - y^2}

⇒x−yx+ya=ax2y2x2−y2⇒a2=x−yx+y×x2y2x2−y2⇒a2=x2y2×(x−y)(x+y)×(x2−y2)⇒a2=x2y2×(x−y)(x+y)×(x−y)(x+y)⇒a2=x2y2(x+y)2⇒a=x2y2(x+y)2⇒a=xy(x+y).\Rightarrow \dfrac{\dfrac{x - y}{x + y}}{a} = \dfrac{a}{\dfrac{x^2y^2}{x^2 - y^2}} \\[1em] \Rightarrow a^2 = \dfrac{x - y}{x + y} \times \dfrac{x^2y^2}{x^2 - y^2} \\[1em] \Rightarrow a^2 = \dfrac{x^2y^2 \times (x - y)}{(x + y) \times (x^2 - y^2)} \\[1em] \Rightarrow a^2 = \dfrac{x^2y^2 \times (x - y)}{(x + y) \times (x - y)(x + y)} \\[1em] \Rightarrow a^2 = \dfrac{x^2y^2}{(x + y)^2} \\[1em] \Rightarrow a = \sqrt{\dfrac{x^2y^2}{(x + y)^2}} \\[1em] \Rightarrow a = \dfrac{xy}{(x + y)}.

Hence, Option 1 is the correct option.

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