The equation of the normal at t of the hyperbola given by x=a2t+1t,y=a2t−1t is

x2−y2=a24t+1t2−t−1t2=a2The normal at x1,y1 is a2xx1+a2yy1=a2+a2=2a2⇒xa2t+1t+ya2t−1t=2⇒xt2+1+yt2−1=at.

The equation of the normal at t of the hyperbola given by x=a2t+1t,y=a2t−1t is

x2−y2=a24t+1t2−t−1t2=a2The normal at x1,y1 is a2xx1+a2yy1=a2+a2=2a2⇒xa2t+1t+ya2t−1t=2⇒xt2+1+yt2−1=at.

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The equation of the normal at t of the hyperbola given by $x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t})$ is

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$\frac{x}{t^{2} - 1} + \frac{y}{t^{2} + 1} = \frac{a}{t}$

$\frac{x}{t^{2} + 1} + \frac{y}{t^{2} - 1} = \frac{a}{t}$

$x (t^{2} + 1) + y (t^{2} - 1) = a t$

$x (t^{2} - 1) + y (t^{2} - 1) = a t$

answer is B.

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Detailed Solution

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

The normal at $(x_{1}, y_{1})$ is $\frac{a^{2} x}{x_{1}} + \frac{a^{2} y}{y_{1}} = a^{2} + a^{2} = 2 a^{2}$

$\Rightarrow \frac{x}{\frac{a}{2} (t + \frac{1}{t})} + \frac{y}{\frac{a}{2} (t - \frac{1}{t})} = 2 \Rightarrow \frac{x}{t^{2} + 1} + \frac{y}{t^{2} - 1} = \frac{a}{t}$ .

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The equation of the normal at t of the hyperbola given by x=a2t+1t,y=a2t−1t is

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The equation of the normal at t of the hyperbola given by $x = \frac{a}{2} (t + \frac{1}{t}), y = \frac{a}{2} (t - \frac{1}{t})$ is