Question

If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$, then $\frac{dy}{dx}$ is equal to:

• A. $\frac{a(1-t^2)}{2t}$
• B. $\frac{a(t^2-1)}{2t}$
• C. $\frac{a(t^2+1)}{2t}$
• D. $\frac{a(t^2-1)}{t}$

Answer 1

Answer: (B)

$\frac{a(t^2-1)}{2t}$

Given, $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2at}{1+t^2}$

On differentiating with respect to $t$, respectively, we get

$\frac{dx}{dt}=\frac{(1+t^2)(0-2t)-(1-t^2)(0+2t)}{{(1+t^2)}^2}$
$=\frac{-4t}{{(1+t^2)}^2}$

and $\frac{dy}{dt}=\frac{(1+t^2)2a-2at\left(2t\right)}{{(1+t^2)}^2}=\frac{2a(1-t^2)}{{(1+t^2)}^2}$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{a(1-t^2)}{-2t}$

$\Rightarrow \frac{dy}{dx}=\frac{a(t^2-1)}{2t}$