Question

Find the equation of the tangent and the normal to the following curve at the indicated points \[x = \frac{2 a t^2}{1 + t^2}, y = \frac{2 a t^3}{1 + t^2}\text { at } t = \frac{1}{2}\] ?

Answer 1

\[x = \frac{2a t^2}{1 + t^2} \text { and y} = \frac{2a t^3}{1 + t^2}\]

\[\frac{dx}{dt} = \frac{\left( 1 + t^2 \right)\left( 4at \right) - 2a t^2 \left( 2t \right)}{\left( 1 + t^2 \right)^2} = \frac{4at}{\left( 1 + t^2 \right)^2}\]

\[ \text { and }\frac{dy}{dt} = \frac{\left( 1 + t^2 \right)6a t^2 - 2a t^3 \left( 2t \right)}{\left( 1 + t^2 \right)^2} = \frac{6a t^2 + 2a t^4}{\left( 1 + t^2 \right)^2}\]

\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{6a t^2 + 2a t^4}{\left( 1 + t^2 \right)^2}}{\frac{4at}{\left( 1 + t^2 \right)^2}} = \frac{6a t^2 + 2a t^4}{4at}\]

\[\text { Slope of tangent,}m= \left( \frac{dy}{dx} \right)_{t = \frac{1}{2}} =\frac{\frac{3a}{2} + \frac{a}{8}}{2a}=\frac{\frac{12a + a}{8}}{2a}=\frac{13a}{8}\times\frac{1}{2a}=\frac{13}{16}\]

\[\text { Now, }\left( x_1 , y_1 \right) = \left( \frac{2a t^2}{1 + t^2}, \frac{2a t^3}{1 + t^2} \right) = \left( \frac{\frac{a}{2}}{1 + \frac{1}{4}}, \frac{\frac{a}{4}}{1 + \frac{1}{4}} \right) = \left( \frac{\frac{a}{2}}{\frac{5}{4}}, \frac{\frac{a}{4}}{\frac{5}{4}} \right) = \left( \frac{2a}{5}, \frac{a}{5} \right)\]

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

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

\[ \Rightarrow y - \frac{a}{5} = \frac{13}{16}\left( x - \frac{2a}{5} \right)\]

\[ \Rightarrow \frac{5y - a}{5} = \frac{13}{16}\left( \frac{5x - 2a}{5} \right)\]

\[ \Rightarrow 5y - a = \frac{13}{16}\left( 5x - 2a \right)\]

\[ \Rightarrow 80y - 16a = 65x - 26a\]

\[ \Rightarrow 65x - 80y - 10a = 0\]

\[ \Rightarrow 13x - 16y - 2a = 0\]

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

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

\[ \Rightarrow y - \frac{a}{5} = \frac{- 16}{13} \left( x - \frac{2a}{5} \right)\]

\[ \Rightarrow \frac{5y - a}{5} = \frac{- 16}{13}\left( \frac{5x - 2a}{5} \right)\]

\[ \Rightarrow 5y - a = \frac{- 16}{13}\left( 5x - 2a \right)\]

\[ \Rightarrow 65y - 13a = - 80x + 32a\]

\[ \Rightarrow 80x + 65y - 45a = 0\]

\[ \Rightarrow 16x + 13y - 9a = 0\]