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A curve $C$ is defined by the equations $$x=\frac{1+t}{1-t}$$ $$y=\frac{1+t^2}{1-t^2}$$ where $t$ is a real parameter.

I found the $\frac{dy}{dx}=\frac{2t}{(t+1)^2}$.

How to prove that $C$ has cartesian equation $y=\frac{x^2+1}{2x}$ by elimating $t$?

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  • $\begingroup$ Can you show first that all your points satisfy $x^2+1/2x = y$? Then you are halfway there. $\endgroup$ – Eric Auld Nov 14 '15 at 3:56
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Apply componendo dividendo to find $$\dfrac{x-1}{x+1}=\dfrac{1+t-(1-t)}{1+t+(1-t)}=t$$

Replace this value of $t$ in $$y=\dfrac{1+t^2}{1-t^2}$$ and simplify

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The points $(x(t),y(t))$ on the curve satisfy $$ x(t)=\frac{1+t}{1-t}$$ and $$y(t)=\frac{1+t^{2}}{1-t^{2}}$$ By squaring the expression for $x(t)$ and adding $1$, we get $$x(t)^{2}+1=\frac{(1+t)^{2}}{(1-t)^{2}}+1=\frac{(1+t)^{2}+(1-t)^{2}}{(1-t)^{2}}=\frac{2+2t^{2}}{(1-t)^{2}}$$ Finally, we divide this expression by $2x(t)=\frac{2(1+t)}{1-t}$, which is the same as multiplying by $\frac{(1-t)}{2(1+t)}$: $$\frac{x(t)^{2}+1}{2x(t)}=\frac{2+2t^{2}}{(1-t)^{2}}\cdot \frac{(1-t)}{2(1+t)}=\frac{1+t^{2}}{(1-t)(1+t)}=\frac{1+t^{2}}{1-t^{2}}$$ is what we get after cancelling $2$ and $(1-t)$ from the numerator and denominator in the second step. Since $y(t)=\frac{1+t^{2}}{1-t^{2}}$, we have that the desired equation for the points on the curve is satisfied: $$y=\frac{x^{2}+1}{2x}$$

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