If x=3at1+t2,y=3at21+t2thendydx=

x=3at1+t2⇒dxdt=1+t23a-3at2t1+t22=3a-3at21+t22=3a1-t21+t22y=3at21+t2⇒dydt=1+t26at-3at22t1+t22=6at1+t22dydx=(dy/dt)(dx/dt)dydx=6at3a1-t2=2t1-t2

If x=3at1+t2,y=3at21+t2thendydx=

x=3at1+t2⇒dxdt=1+t23a-3at2t1+t22=3a-3at21+t22=3a1-t21+t22y=3at21+t2⇒dydt=1+t26at-3at22t1+t22=6at1+t22dydx=(dy/dt)(dx/dt)dydx=6at3a1-t2=2t1-t2

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If $x = \frac{3 a t}{1 + t^{2}}, y = \frac{3 a t^{2}}{1 + t^{2}} then \frac{d y}{d x} =$

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

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

$2 t (t^{2} - 1)$

$- 2 t (t^{2} - 1)$

answer is A.

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Detailed Solution

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

$y = \frac{3 a t^{2}}{1 + t^{2}} \Rightarrow \frac{d y}{d t} = \frac{(1 + t^{2}) 6 a t - 3 a t^{2} (2 t)}{{(1 + t^{2})}^{2}} = \frac{6 a t}{{(1 + t^{2})}^{2}}$