Question

Find the equation of the tangent and the normal to the following curve at the indicated point\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ at }\left( a\cos\theta, b\sin\theta \right)\] ?

Answer 1

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

\[\text { Differentiating both sides w.r.t. x }, \]

\[ \Rightarrow \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0\]

\[ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx} = \frac{- 2x}{a^2}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- x b^2}{y a^2}\]

\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_\left( a \cos \theta, b \sin \theta \right) =\frac{- a \cos \theta \left( b^2 \right)}{b \sin \theta \left( a^2 \right)}=\frac{- b \cos \theta}{a \sin \theta}\]

\[\text { Given } \left( x_1 , y_1 \right) = \left( a \cos \theta, b \sin \theta \right)\]

\[\text { Equation of tangent is },\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - b \sin \theta = \frac{- b \cos \theta}{a \sin \theta}\left( x - a \cos \theta \right)\]

\[ \Rightarrow ay \sin \theta - \text { ab }\sin^2 \theta = - bx \cos \theta + ab \cos^2 \theta\]

\[ \Rightarrow bx \cos \theta + ay \sin \theta = ab\]

\[\text{ Dividing by ab},\]

\[ \Rightarrow \frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1\]

\[\text { Equation of normal is} ,\]

\[y - y_1 = \frac{- 1}{m} \left( x - x_1 \right)\]

\[ \Rightarrow y - b \sin \theta = \frac{a \sin \theta}{b \cos \theta}\left( x - a \cos \theta \right)\]

\[ \Rightarrow by \cos \theta - b^2 \sin \theta \cos \theta = ax \sin \theta - a^2 \sin \theta \cos \theta\]

\[ \Rightarrow ax \sin \theta - by \cos \theta = \left( a^2 - b^2 \right)\sin \theta \cos \theta\]

\[\text { Dividing by }\sin \theta \cos \theta, \]

\[ax \sec \theta - \text { by }\ cosec \theta = \left( a^2 - b^2 \right)\]