Question

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

• A. $\frac{4}{3}$
• B. $-\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $-\frac{3}{4}$

Answer 1

The correct option is A $\frac{4}{3}$

Given that
$x=\frac{2t}{1+t^2}$, $y=\frac{1-t^2}{1+t^2}$
By differentiating w.r. to $t$, we get
$\frac{dx}{dt}=\frac{\left((1+t^2)\frac{d}{dt}\left(2t\right)-2t\frac{d}{dt}(1+t^2)\right)}{{(1+t^2)}^2}$
And $\frac{dy}{dt}=\frac{\left((1+t^2)\frac{d}{dt}(1-t^2)-(1-t^2)\frac{d}{dt}(1+t^2)\right)}{{(1+t^2)}^2}$
$\frac{dx}{dt}=\frac{(1+t^2)2-2t\times 2t}{{(1+t^2)}^2}=\frac{2-2t^2}{{(1+t^2)}^2}$

$\frac{dy}{dt}=\frac{(1+t^2)(-2t)-(1-t^2)2t}{{(1+t^2)}^2}=\frac{-4t}{{(1+t^2)}^2}$
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-4t}{2-2t^2}=\frac{2t}{t^2-1}$
$\therefore {\left(\frac{dy}{dx}\right)}_{t=2}=\frac{4}{3}$