Question

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

Answer 1

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

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

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