Question

If x = `(4t)/(1 + t^2),  y = 3((1 - t^2)/(1 + t^2))` then show that `dy/dx = (-9x)/(4y)`.

Answer 1

x = `(4t)/(1 + t^2)`

Differentiating both sides w.r.t. ‘t’, we get

`dx/dt = ((1 + t^2)*d/dx (4t) - 4t * d/dx (1 + t^2))/(1 + t^2)^2`

`= ((1 + t^2)(4) - 4t(0 + 2t))/(1 + t^2)^2`

`= (4 + 4t^2 - 8t^2)/(1 + t^2)^2`

`= (4 - 4t^2)/(1 + t^2)^2`

`= (4(1 - t^2))/(1 + t^2)^2`

y = `3((1 - t^2)/(1 + t^2))`

Differentiating both sides w.r.t. ‘t’, we get

`dy/dt = 3 d/dx ((1 - t^2)/(1 + t^2))`

`= 3 [((1 + t^2) d/dt (1 - t^2) - (1 - t^2) * d/dt (1 + t^2))/(1 + t^2)^2]`

`= 3[((1 + t^2)(0 - 2t) - (1 - t^2)(0 + 2t))/(1 + t^2)^2]`

`= 3 [(-2t (1 + t^2) - 2t(1 - t^2))/(1 + t^2)^2]`

`= 3(- 2t) [(1 + t^2 + 1 - t^2)/(1 + t^2)^2]`

`= - 6t xx 2/(1 + t^2)^2`

`= (- 12t)/(1 + t^2)^2`

∴ `dy/dx = ((dy/dt))/((dx/dt)) = ((-12t)/(1 + t^2)^2)/((4(1 - t^2))/(1 + t^2)^2)`

∴ `dy/dx = (- 3t)/(1 - t^2)`   ....(i)

Also `(- 9x)/(4y) = (- 9((4t)/(1 + t^2)))/(4xx3 ((1 - t^2)/(1 + t^2))) = (- 3t)/(1 - t^2)`     ....(ii)

From (i) and (ii), we get

`dy/dx = (- 9x)/(4y)`