# Solve this

Question:

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

Solution:

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

$\Rightarrow \frac{d x}{d t}=a\left[\frac{\left(1-t^{2}\right) \frac{d}{d t}\left(1+t^{2}\right)-\left(1+t^{2}\right) \frac{d}{d t}\left(1-t^{2}\right)}{\left(1-t^{2}\right)^{2}}\right]$       [Using quotient rule]

$\Rightarrow \frac{d x}{d t}=a\left[\frac{\left(1-t^{2}\right)(2 t)-\left(1+t^{2}\right)(-2 t)}{\left(1-t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=a\left[\frac{2 t-2 t^{3}+2 t+2 t^{3}}{\left(1-t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d x}{d t}=\frac{4 a t}{\left(1-t^{2}\right)^{2}}$                            .....(1)

and,

$y=\frac{2 t}{1-t^{2}}$

$\Rightarrow \frac{d y}{d t}=2\left[\frac{\left(1-t^{2}\right) \frac{d}{d t}(t)-t \frac{d}{d t}\left(1-t^{2}\right)}{\left(1-t^{2}\right)^{2}}\right]$    [Using quotient rule]

$\Rightarrow \frac{d y}{d t}=2\left[\frac{\left(1-t^{2}\right)(1)-t(-2 t)}{\left(1-t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d y}{d t}=2\left[\frac{1-t^{2}+2 t^{2}}{\left(1-t^{2}\right)^{2}}\right]$

$\Rightarrow \frac{d y}{d t}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}}$               .....(2)

Dividing equation (ii) by (i),

$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2\left(1+t^{2}\right)}{\left(1-t^{2}\right)^{2}} \times \frac{\left(1-t^{2}\right)^{2}}{4 a t}$

$\Rightarrow \frac{d y}{d x}=\frac{\left(1+t^{2}\right)}{2 a t}$