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प्रश्न
Find the equation of the tangent and the normal to the following curve at the indicated point\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ at }\left( a\cos\theta, b\sin\theta \right)\] ?
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उत्तर
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
\[\text { Differentiating both sides w.r.t. x }, \]
\[ \Rightarrow \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0\]
\[ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = \frac{- 2x}{a^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- x b^2}{y a^2}\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( a \cos \theta, b \sin \theta \right) =\frac{- a \cos \theta \left( b^2 \right)}{b \sin \theta \left( a^2 \right)}=\frac{- b \cos \theta}{a \sin \theta}\]
\[\text { Given } \left( x_1 , y_1 \right) = \left( a \cos \theta, b \sin \theta \right)\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - b \sin \theta = \frac{- b \cos \theta}{a \sin \theta}\left( x - a \cos \theta \right)\]
\[ \Rightarrow ay \sin \theta - \text { ab }\sin^2 \theta = - bx \cos \theta + ab \cos^2 \theta\]
\[ \Rightarrow bx \cos \theta + ay \sin \theta = ab\]
\[\text{ Dividing by ab},\]
\[ \Rightarrow \frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1\]
\[\text { Equation of normal is} ,\]
\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]
\[ \Rightarrow y - b \sin \theta = \frac{a \sin \theta}{b \cos \theta}\left( x - a \cos \theta \right)\]
\[ \Rightarrow by \cos \theta - b^2 \sin \theta \cos \theta = ax \sin \theta - a^2 \sin \theta \cos \theta\]
\[ \Rightarrow ax \sin \theta - by \cos \theta = \left( a^2 - b^2 \right)\sin \theta \cos \theta\]
\[\text { Dividing by }\sin \theta \cos \theta, \]
\[ax \sec \theta - \text { by }\ cosec \theta = \left( a^2 - b^2 \right)\]
