Question:

If, for non-zero real variables x,yx, y and a real parameter a>1a>1, x:y=(a+1):(a1)x : y = (a+1) : (a-1). Then, the ratio (x2y2):(x2+y2)(x^2 - y^2) : (x^2 + y^2) is:

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For ratios involving squares in fractions, use substitution and simplifications effectively to simplify the terms systematically.
Updated On: Jan 23, 2025
  • 2a:(a2+1)2a : (a^2+1)
  • a:(a2+1)a : (a^2+1)
  • 2a:(a21)2a : (a^2-1)
  • a:(a21)a : (a^2-1)
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The Correct Option is A

Solution and Explanation

Step 1: From the given condition, x:y=(a+1):(a1)x : y = (a+1) : (a-1), we write: xy=a+1a1. \frac{x}{y} = \frac{a+1}{a-1}. Step 2: Substitute this into the ratio (x2y2):(x2+y2)(x^2 - y^2) : (x^2 + y^2): x2y2x2+y2=(xy)21(xy)2+1. \frac{x^2 - y^2}{x^2 + y^2} = \frac{\left(\frac{x}{y}\right)^2 - 1}{\left(\frac{x}{y}\right)^2 + 1}. Step 3: Replace xy\frac{x}{y} with a+1a1\frac{a+1}{a-1}: (a+1a1)21(a+1a1)2+1=(a+1)2(a1)21(a+1)2(a1)2+1. \frac{\left(\frac{a+1}{a-1}\right)^2 - 1}{\left(\frac{a+1}{a-1}\right)^2 + 1} = \frac{\frac{(a+1)^2}{(a-1)^2} - 1}{\frac{(a+1)^2}{(a-1)^2} + 1}. Simplify: (a+1)2(a1)2(a1)2(a+1)2+(a1)2(a1)2=(a+1)2(a1)2(a+1)2+(a1)2. \frac{\frac{(a+1)^2 - (a-1)^2}{(a-1)^2}}{\frac{(a+1)^2 + (a-1)^2}{(a-1)^2}} = \frac{(a+1)^2 - (a-1)^2}{(a+1)^2 + (a-1)^2}. Step 4: Expand and simplify: (a+1)2(a1)2=4a,(a+1)2+(a1)2=2a2+2. (a+1)^2 - (a-1)^2 = 4a, \quad (a+1)^2 + (a-1)^2 = 2a^2 + 2. So the ratio becomes: 4a2a2+2=2aa2+1. \frac{4a}{2a^2 + 2} = \frac{2a}{a^2+1}. Step 5: The required ratio is: 2a:(a2+1). 2a : (a^2 + 1).
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