Find the Cartesian equation of the curve given by the parametric equations: x=(1-t^2)/(1+t^2) and y=2t/(1+t^2). I know how to do Cartesian equations, but I am unsure how to solve for t in this one.

Question

anonymous · Answer

$$\begin{align*}\cos y&=\cos\left(2\times\frac{y}{2}\right)\\
&=\cos^2\frac{y}{2}-\sin^2\frac{y}{2}\\
\sin y&=\sin\left(2\times\frac{y}{2}\right)\\
&=2\sin\frac{y}{2}\cos\frac{y}{2}\end{align*}$$
Recall that $\cos^2y+\sin^2y=1$, so we should have
$$1=\left(\cos^2\frac{y}{2}-\sin^2\frac{y}{2}\right)^2+\left(2\sin\frac{y}{2}\cos\frac{y}{2}\right)^2$$
Suppose we have
$$\cos\frac{y}{2}=\frac{1}{\sqrt{1+t^2}}~~~~\text{and}~~~~\sin\frac{y}{2}=\frac{t}{\sqrt{1+t^2}}$$
Then indeed, we have
$$\begin{align*}1&=\left(\cos^2\frac{y}{2}-\sin^2\frac{y}{2}\right)^2+\left(2\sin\frac{y}{2}\cos\frac{y}{2}\right)^2\\
&=\left(\left(\frac{1}{\sqrt{1+t^2}}\right)^2-\left(\frac{t}{\sqrt{1+t^2}}\right)^2\right)^2+\left(2\times\frac{1}{\sqrt{1+t^2}}\times\frac{t}{\sqrt{1+t^2}}\right)^2\\
&=\left(\frac{1}{1+t^2}-\frac{t^2}{1+t^2}\right)^2+\left(\frac{2t}{1+t^2}\right)^2\\
&=\left(\frac{1-t^2}{1+t^2}\right)^2+\left(\frac{2t}{1+t^2}\right)^2\\
1&=x^2+y^2
\end{align*}$$
Look familiar?