The locus represented by x=a2t+1t,y=a2t-1t is

Given x=a2(t+1t), y=a2(t-1t) ⇒2xa=t+1t-①, 2ya=t-1t-② ①2 - ②2 ⇒4x2a2-4y2a2=t+1t2-t-1t2 ⇒4x2a2-y2a2=4 ⇒x2-y2=a2

The locus represented by x=a2t+1t,y=a2t-1t is

Given x=a2(t+1t), y=a2(t-1t) ⇒2xa=t+1t-①, 2ya=t-1t-② ①2 - ②2 ⇒4x2a2-4y2a2=t+1t2-t-1t2 ⇒4x2a2-y2a2=4 ⇒x2-y2=a2

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The locus represented by
$x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t})$ is

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

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

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

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

answer is B.

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Detailed Solution

$Given x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t}) \Rightarrow \frac{2 x}{a} = (t + \frac{1}{t}) - ①, \frac{2 y}{a} = t - \frac{1}{t} - ② ①^{2} - ②^{2} \Rightarrow \frac{4 x^{2}}{a^{2}} - \frac{4 y^{2}}{a^{2}} = {(t + \frac{1}{t})}^{2} - {(t - \frac{1}{t})}^{2} \Rightarrow 4 [\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}}] = 4 \Rightarrow x^{2} - y^{2} = a^{2}$