Question

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

\[ \Rightarrow bx \cos \theta - \text { ay } \sin \theta \cos \theta = ab \left( 1 - \sin^2 \theta \right)\]

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

\[\text { Dividing by ab } \cos^2 \theta,\]

\[ \Rightarrow \frac{x}{a}\sec \theta - \frac{y}{b}\tan \theta = 1\]

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

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

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

\[ \Rightarrow yb - b^2 \tan \theta = - ax \sin \theta + a^2 \tan\theta\]

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

\[\text { Dividing by tan } \theta, \]

\[ax \cos \theta + by \cot \theta = \left( a^2 + b^2 \right)\]