\[\begin{array}{1 1 }(1)3\theta=27t-80&(2)5\theta=27t-80\\(3)3\theta=27t+80&(4)\theta=\frac{1}{t}\end{array}\]

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$\theta =\large\frac{1}{t}$

$\large\frac{d \theta}{dt}=\large\frac{-1}{t^2}$

$\large\frac{d \theta}{dt} (-3,\large\frac{-1}{3} ) =\large\frac{-1}{3^2}=\frac{-1}{9}$

Slope of the tangent $=-\large\frac{1}{9}$

$\therefore$ Slope of the normal $= -\large\frac{1}{\Large\frac{-1}{9}}$$=9$

$\theta + \large\frac{1}{3}$$=9(t+3)$

$3 \theta +1 =27 (t+3)$

$3 \theta +1=27t +8t$

$3 \theta =27 t +80$

Hence 3 is the correct answer.

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