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