If normal to the rectangular hyperbola $x y = 4$ at the point $t_{1} (= 4)$ meets the curve again at the point $t_{2}$ , then $\frac{1}{| t_{2} |} =$

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Solution

In the rectangular hyperbola $x y = 4$ , equation of normal at $t_{1}$ is
$(y - \frac{2}{t_{1}}) = t_{1}^{2} (x - 2 t_{1}) \Rightarrow y - \frac{1}{2} = 16 (x - 8) \dots (1)$
Here $(1)$ passes through $(2 t_{2}, \frac{2}{t_{2}})$
$\Rightarrow (\frac{2}{t_{2}} - \frac{1}{2}) = 16 (2 t_{2} - 8)$
$\Rightarrow \frac{4 - t_{2}}{2 t_{2}} = 32 (t_{2} - 4)$
$\Rightarrow t_{2} = - \frac{1}{64} [∵ t_{1} \neq t_{2}]$

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If normal to the rectangular hyperbola xy=4 at the point t1(=4) meets the curve again at the point t2, then 1|t2|=

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