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If a normal makes an angle $\theta$ with positive $x$-axis then the slope of the curve at the point where the normal is drawn is

\[\begin{array}{1 1}(1)-\cot\theta&(2)\tan\theta\\(3)-\tan\theta&(4)\cot\theta\end{array}\]

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Given the normal makes a angle $\theta$ with x-axis
Slope of the normal $= \tan \theta$
Slope of the tangent at the point where the normal is draw $=-\large\frac{1}{\tan \theta}$
$\qquad= -\cot \theta$
Hence 1 is the correct answer.
answered May 19, 2014 by meena.p
 

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