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a
2a units from origin
b
a units from origin
c
$\frac{1}{2}\mathrm{a}$ units from origin
d
5a units from origin

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detailed solution

Correct option is B

$\begin{array}{l}\mathrm{x}=\mathrm{acos}\mathrm{\theta }+\mathrm{a\theta sin}\mathrm{\theta },\mathrm{y}=\mathrm{asin}\mathrm{\theta }-\mathrm{a\theta cos}\mathrm{\theta }\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a\theta sin}\mathrm{\theta }}{\mathrm{a\theta cos}\mathrm{\theta }}=\frac{\mathrm{sin}\mathrm{\theta }}{\mathrm{cos}\mathrm{\theta }}\end{array}$

Slope of normal $=\frac{-\mathrm{cos}\mathrm{\theta }}{\mathrm{sin}\mathrm{\theta }}$

Equation of normal at $\mathrm{\theta }$

$\begin{array}{l}\mathrm{y}-\left[\mathrm{asin}\mathrm{\theta }-\mathrm{a\theta cos}\mathrm{\theta }\right]=\frac{-\mathrm{cos}\mathrm{\theta }}{\mathrm{sin}\mathrm{\theta }}\left[\mathrm{x}-\left(\mathrm{acos}\mathrm{\theta }+\mathrm{a\theta sin}\mathrm{\theta }\right)\right]\\ \mathrm{ysin}\mathrm{\theta }-{\mathrm{asin}}^{2}\mathrm{\theta }+\mathrm{a\theta sin}\mathrm{\theta cos}\mathrm{\theta }=-\mathrm{xcos}\mathrm{\theta }+{\mathrm{acos}}^{2}\mathrm{\theta }+\mathrm{a\theta sin}\mathrm{\theta cos}\mathrm{\theta }\\ \mathrm{xcos}\mathrm{\theta }+\mathrm{ysin}\mathrm{\theta }-\mathrm{a}=0\end{array}$

The distance from origin to the normal

$=\frac{|-\mathrm{a}|}{\sqrt{{\mathrm{cos}}^{2}\mathrm{\theta }+{\mathrm{sin}}^{2}\mathrm{\theta }}}=\mathrm{a}$

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