$Given P = (1, 0) and Q = (- 1, 0) and R is a variable point on one side of the line PQ such that$
$∠ R P Q - ∠ R Q P = \frac{π}{4} . The locus of the point R is$

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$x^{2} + y^{2} + 2 x y = 1$

$x^{2} - y^{2} + 2 x y = 1$

$x^{2} + y^{2} - 2 x y = 1$

$x^{2} - y^{2} - 2 x y = 1$

answer is D.

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Detailed Solution

$α - β = \frac{π}{4}, Tan (α - β) = Tan \frac{π}{4} \Rightarrow \frac{Tan α - Tan β}{1 + Tan α Tan β} = 1$

$\tan β = \frac{y}{1 + x}, \tan α = \frac{y}{1 - x}$

$\frac{y}{1 - x} - \frac{y}{1 + x} = 1 + (\frac{y}{1 - x}) (\frac{y}{1 + x})$

$y (1 + x - 1 + x) = 1 - x^{2} + y^{2} \Rightarrow x^{2} - y^{2} + 2 x y = 1$

$Req. locus is x^{2} - y^{2} + 2 x y = 1$

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