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प्रश्न
Find the equation of the tangent and the normal to the following curve at the indicated points x = a(θ + sinθ), y = a(1 − cosθ) at θ ?
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उत्तर
\[x = a\left( \theta + \sin\theta \right) \text { and }y = a\left( 1 - \cos\theta \right)\]
\[\frac{dx}{d\theta} = a\left( 1 + \cos\theta \right) \text { and } \frac{dy}{d\theta} = a\sin\theta\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\sin\theta}{a\left( 1 + \cos\theta \right)} = \frac{\sin\theta}{\left( 1 + \cos\theta \right)} = \frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} = \tan\frac{\theta}{2} . . . \left( 1 \right)\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\theta =\tan\frac{\theta}{2}\]
\[\text { Now }, \left( x_1 , y_1 \right) = \left[ a\left( \theta + \sin\theta \right), a\left( 1 - \cos\theta \right) \right] \]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - a\left( 1 - \cos\theta \right) = \tan\frac{\theta}{2}\left[ x - a\left( \theta + \sin\theta \right) \right]\]
\[ \Rightarrow y - a\left( 2 \sin^2 \frac{\theta}{2} \right) = x\tan\frac{\theta}{2} - a\theta\tan\frac{\theta}{2} - a\tan\frac{\theta}{2}\sin\theta\]
\[ \Rightarrow y - a\left( 2 \sin^2 \frac{\theta}{2} \right) = x\tan\frac{\theta}{2} - a\theta\tan\frac{\theta}{2} - a\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}2\sin\frac{\theta}{2}\cos\frac{\theta}{2}........... [From (1)]\]
\[ \Rightarrow y - 2a \sin^2 \frac{\theta}{2} = \left( x - a\theta \right)\tan\frac{\theta}{2} - 2a \sin^2 \frac{\theta}{2}\]
\[ \Rightarrow y = \left( x - a\theta \right)\tan\frac{\theta}{2}\]
\[\text { Equation of normal is },\]
\[y - a\left( 1 - \cos\theta \right) = - \cot\frac{\theta}{2}\left[ x - a\left( \theta + \sin\theta \right) \right]\]
\[ \Rightarrow \tan \frac{\theta}{2}\left[ y - a\left( 2 \sin^2 \frac{\theta}{2} \right) \right] = - x + a\theta + a\sin\theta\]
\[ \Rightarrow \tan \frac{\theta}{2}\left[ y - a\left\{ 2 \left( 1 - \cos^2 \frac{\theta}{2} \right) \right\} \right] = - x + a\theta + a\sin\theta\]
\[ \Rightarrow \tan \frac{\theta}{2}\left( y - 2a \right) + a \left( 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \right) = - x + a\theta + asin\theta\]
\[ \Rightarrow \tan \frac{\theta}{2}\left( y - 2a \right) + a\sin\theta = - x + a\theta + asin\theta\]
\[ \Rightarrow \tan \frac{\theta}{2}\left( y - 2a \right) = - x + a\theta\]
\[ \Rightarrow \tan \frac{\theta}{2}\left( y - 2a \right) + x - a\theta = 0\]
