A line is drawn through the point $P (- 1, 1)$ to meet the curve $y = \frac{1}{x}$ in the points A and B (Points A and B lie on same side of P). A point R is chosen on this line such that PA, PR and PB are in Arithmetic Progression. Then the locus of R may be

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$x - y = 2$

$2 x y = x - y$

$2 x y = 1$

$2 x y = x + y$

answer is B.

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Detailed Solution

Any point on the line through $P (- 1, 1)$ at a distance of $r$ from P will be $(- 1 + r \cos θ, 1 + r \sin θ)$ , where $\tan θ$ is slope of the line.
If this line meets $x y = 1$ in A & B, then $(- 1 + r \cos θ) (1 + r \sin θ) = 1$ will have PA & PB as its roots.
$r^{2} \sin θ \cos θ + r (\cos θ - \sin θ) - 2 = 0$
Also, for the point R, $x = - 1 + P R \cos θ & y = 1 + P R \sin θ$
Hence $\cos θ = \frac{x + 1}{P R} & \sin θ = \frac{y - 1}{P R}$
Now as PA, PR & PB are in A.P., hence PA+PB=2PR
$\Rightarrow \frac{\sin θ - \cos θ}{\sin θ \cos θ} = 2 P R \Rightarrow \frac{\frac{y - 1}{P R} - \frac{x + 1}{P R}}{\frac{x + 1}{P R} . \frac{y - 1}{P R}} = 2 P R \Rightarrow 2 x y = x - y$