Question

Find the equation of the tangent and the normal to the following curve at the indicated points  x = asect, y = btant at t ?

Answer 1

\[x = a \sec t \text{ and }y = b \tan t\]

\[\frac{dx}{dt} = a \sec t \tan t \text { and } \frac{dy}{dt} = b \sec^2 t\]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b \sec^2 t}{a \sec t \tan t} = \frac{b}{a}\ cosec\ t\]

\[\text { Slope of tangent, }m= \left( \frac{dy}{dx} \right)_{t = t} =\frac{b}{a}\ cosec\ t\]

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

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

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

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

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

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

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

\[ \Rightarrow ay \sin t \cos t - ab \sin^2 t = bx \cos t - ab\]

\[ \Rightarrow bx \cos t - ay \sin t \cos t - ab\left( 1 - \sin^2 t \right) = 0\]

\[ \Rightarrow bx \cos t - ay \sin t \cos t = ab \cos^2 t\]

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

\[bx \sec t - ay \tan t = ab\]

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

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

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

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

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

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

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

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

\[\text { Dividing both sides by sint },\]

\[ax \cos t + by \cot t = a^2 + b^2\]