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The equation of the normal to the curve $\theta =\large\frac{1}{t}$ at the point $(-3 , -1/3)$ is

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

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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.
answered May 19, 2014 by meena.p
 

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