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प्रश्न
Find the equation of the tangent and the normal to the following curve at the indicated points x = θ + sinθ, y = 1 + cosθ at θ = \[\frac{\pi}{2}\] ?
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उत्तर
\[x = \theta + \sin\theta \text { and }y = 1 + \cos\theta\]
\[\frac{dx}{d\theta} = 1 + \cos\theta \text { and } \frac{dy}{d\theta} = - \sin\theta\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{- \sin\theta}{1 + \cos\theta}\]
\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_{\theta = \frac{\pi}{2}} =\frac{- \sin\frac{\pi}{2}}{1 + \cos\frac{\pi}{2}}=\frac{- 1}{1 + 0}=-1\]
\[\text { Now,} \left( x_1 , y_1 \right) = \left( \frac{\pi}{2} + \sin\frac{\pi}{2}, 1 + \cos\frac{\pi}{2} \right) = \left( \frac{\pi}{2} + 1, 1 \right)\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 1 = - 1\left( x - \frac{\pi}{2} - 1 \right)\]
\[ \Rightarrow 2y - 2 = - 2x + \pi + 2\]
\[ \Rightarrow 2x + 2y - \pi - 4 = 0\]
\[\text { Equation of normal is },\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - 1 = 1 \left( x - \frac{\pi}{2} - 1 \right)\]
\[ \Rightarrow 2y - 2 = 2x - \pi - 2\]
\[ \Rightarrow 2x - 2y = \pi\]
