Question

If \[x = a \left( \frac{1 + t^2}{1 - t^2} \right) \text { and y } = \frac{2t}{1 - t^2}, \text { find } \frac{dy}{dx}\] ?

Answer 1

\[\text { We have, x } = a\left( \frac{1 + t^2}{1 - t^2} \right)\]

\[\Rightarrow \frac{dx}{dt} = a\left[ \frac{\left( 1 - t^2 \right)\frac{d}{dt}\left( 1 + t^2 \right) - \left( 1 + t^2 \right)\frac{d}{dt}\left( 1 - t^2 \right)}{\left( 1 - t^2 \right)^2} \right] \left[ \text { Using quotient rule } \right]\]
\[ \Rightarrow \frac{dx}{dt} = a\left[ \frac{\left( 1 - t^2 \right)\left( 2t \right) - \left( 1 + t^2 \right)\left( - 2t \right)}{\left( 1 - t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = a\left[ \frac{2t - 2 t^3 + 2t + 2 t^3}{\left( 1 - t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dx}{dt} = \frac{4at}{\left( 1 - t^2 \right)^2} . . . \left( i \right)\]
\[\text { and,} \]
\[ y = \frac{2t}{1 - t^2}\]

\[\Rightarrow \frac{dy}{dt} = 2\left[ \frac{\left( 1 - t^2 \right)\frac{d}{dt}\left( t \right) - t\frac{d}{dt}\left( 1 - t^2 \right)}{\left( 1 - t^2 \right)^2} \right] \left[ \text { Using quotient rule } \right]\]
\[ \Rightarrow \frac{dy}{dt} = 2\left[ \frac{\left( 1 - t^2 \right)\left( 1 \right) - t\left( - 2t \right)}{\left( 1 - t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dt} = 2\left[ \frac{1 - t^2 + 2 t^2}{\left( 1 - t^2 \right)^2} \right]\]
\[ \Rightarrow \frac{dy}{dt} = \frac{2\left( 1 + t^2 \right)}{\left( 1 - t^2 \right)^2} . . . \left( ii \right)\]
\[\text { Dividing equation } \left( ii \right) \text { by } \left( i \right), \]
\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2\left( 1 + t^2 \right)}{\left( 1 - t^2 \right)^2} \times \frac{\left( 1 - t^2 \right)^2}{4at}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( 1 + t^2 \right)}{2at}\]