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$x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 a t}{1+t^2}$, then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{a\left(t^2-1\right)}{2 t}$
$x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 a t}{1+t^2}$
Differentiating with respect to $t$, we get
$\begin{aligned} & \frac{d x}{d t}=\frac{\left(1+t^2\right)(0-2 t)-\left(1-t^2\right)(0+2 t)}{\left(1+t^2\right)^2}=-\frac{4 t}{\left(1+t^2\right)^2} \\ & \text { and } \frac{d y}{d t}=\frac{\left(1+t^2\right) 2 a-2 a t(2 t)}{\left(1+t^2\right)^2}=\frac{2 a\left(1-t^2\right)}{\left(1+t^2\right)^2} \\ & \Rightarrow \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a\left(1-t^2\right)}{-2 t} ; \therefore \frac{d y}{d x}=\frac{a\left(t^2-1\right)}{2 t} .\end{aligned}$
Differentiating with respect to $t$, we get
$\begin{aligned} & \frac{d x}{d t}=\frac{\left(1+t^2\right)(0-2 t)-\left(1-t^2\right)(0+2 t)}{\left(1+t^2\right)^2}=-\frac{4 t}{\left(1+t^2\right)^2} \\ & \text { and } \frac{d y}{d t}=\frac{\left(1+t^2\right) 2 a-2 a t(2 t)}{\left(1+t^2\right)^2}=\frac{2 a\left(1-t^2\right)}{\left(1+t^2\right)^2} \\ & \Rightarrow \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a\left(1-t^2\right)}{-2 t} ; \therefore \frac{d y}{d x}=\frac{a\left(t^2-1\right)}{2 t} .\end{aligned}$
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