Q.If $x=\dfrac{2bt}{1+t^2}$, $y=a\dfrac{1-t^2}{1+t^2}$, show that $\dfrac{dx}{dy}=-\dfrac{b^2y}{a^2x}$.

Sometimes both $x$ and $y$ are given not directly in terms of each other but through a third variable $t$ (the parameter): $x=f(t)$, $y=g(t)$. For example $x=a\cos t$, $y=a\sin t$ traces a circle $x^2+y^2=a^2$ as $t$ varies, without $y$ ever being written explicitly as a function of $x$. When $x$ and $y$ are both differentiable functions of the parameter $t$ with $\dfrac{dx}{dt}\ne0$, then $y$ is a differentiable function of $x$, and $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ — found by differentiating $y$ and $x$ separately with respect to $t$ and dividing. This follows from the increment definition: $\dfrac{\delta y}{\delta x}=\dfrac{\delta y/\delta t}{\delta x/\delta t}$, and taking …