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If $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{a\left(t^{2}-1\right)}{2 t}$
Given, $x=\frac{1-t^{2}}{1+t^{2}}$ and $y=\frac{2 a t}{1+t^{2}}$
On differentiating w.r.t. $t$, respectively, we get
$$
\begin{aligned}
\frac{d x}{d t} &=\frac{\left(1+t^{2}\right)(0-2 t)-\left(1-t^{2}\right)(0+2 t)}{\left(1+t^{2}\right)^{2}} \\
&=\frac{-4 t}{\left(1+t^{2}\right)^{2}}
\end{aligned}
$$
and $\frac{d y}{d t}=\frac{\left(1+t^{2}\right) 2 a-2 a t(2 t)}{\left(1+t^{2}\right)^{2}}=\frac{2 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a\left(1-t^{2}\right)}{-2 t}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{a\left(t^{2}-1\right)}{2 t}$
On differentiating w.r.t. $t$, respectively, we get
$$
\begin{aligned}
\frac{d x}{d t} &=\frac{\left(1+t^{2}\right)(0-2 t)-\left(1-t^{2}\right)(0+2 t)}{\left(1+t^{2}\right)^{2}} \\
&=\frac{-4 t}{\left(1+t^{2}\right)^{2}}
\end{aligned}
$$
and $\frac{d y}{d t}=\frac{\left(1+t^{2}\right) 2 a-2 a t(2 t)}{\left(1+t^{2}\right)^{2}}=\frac{2 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a\left(1-t^{2}\right)}{-2 t}$
$\Rightarrow \quad \frac{d y}{d x}=\frac{a\left(t^{2}-1\right)}{2 t}$
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