The position of a particle moving in the \(xy\)-plane is given by the parametric equations \(x(t)=\dfrac{6t}{t+1}\) and \(y(t)=-\dfrac{8}{t^2+4}\). What is the slope of the line tangent to the path of the particle at the point where \(t=2\) ?

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To find the slope of the line tangent, we need to find:

$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$

Take the derivative of \(x(t)\) and \(y(t)\) using the quotient rule. We obtain:

$$ \frac{dx}{dt}=x'(t)=\frac{6}{(t+1)^2} $$ $$ \frac{dy}{dt}=y'(t)=\frac{16t}{(t^2+4)^2} $$

At \(t=2\),

$$ x'(2) = \frac{2}{3} $$ $$ y'(2) = \frac{1}{2} $$ $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \boxed{\frac{3}{4}} $$