If `x=(1-t^(2))/(1+t^(2))` and `y=(2t)/(1+t^(2))`, then `(dy)/(dx)` is equal to

Options

A. `(a(1-t^2))/(2t)`
B. `(a(t^(2)-1))/(2t)`
C. `(a(t^(2)+1))/(2t)`
D. `(a(t^(2)-1))/(t)`

Answer

B

Explanation

Given, `x=(1-t^(2))/(1+t^(2)) and y=(2at)/(1+t^(2))`
On differentiating w.r.t. respectively, we get
`(dx)/(dt)=((1+t^(2))(0-2t)-(1-t^(2))(0+2t))/((1+t^(2))^(2))`
`=(-4t)/((1+t^(2))^(2))`
and `(dy)/(dt)=((1+t^(2))2a-2at(2t))/((1+t^(2))^(2))=(2a(1-t^(2)))/((1+t^(2))^(2))`
`:. (dy)/(dx)=(dy//dt)/(dx//dt)=(a(1-t^(2)))/(-2t)`
`rArr (dy)/(dx)=(a(t^(2)-1))/(2t)`