Question

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

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

Answer 1

The correct option is A $\frac{2t}{1-t^2}$

Given,

$x=\frac{3at}{1+t^2}$

$\frac{dx}{dt}=3a\frac{\frac{d}{dt}\left(t\right)(1+t^2)-\frac{d}{dt}(1+t^2)t}{{(1+t^2)}^2}$

$\therefore \frac{dx}{dt}=\frac{3a(1-t^2)}{{(1+t^2)}^2}$

$y=\frac{3a{t}^2}{1+t^2}$

$\frac{dy}{dt}=3a\frac{\frac{d}{dt}\left(t^2\right)(1+t^2)-\frac{d}{dt}(1+t^2)t^2}{{(1+t^2)}^2}$

$\therefore \frac{dy}{dt}=\frac{6at}{{(1+t^2)}^2}$

Now,

$\frac{dy}{dx}=\frac{\frac{6at}{{(1+t^2)}^2}}{\frac{3a(1-t^2)}{{(1+t^2)}^2}}$

$=\frac{6at}{3a(1-t^2)}$

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