Prove that the evolute of the tractrix $x=a(\cos t+\log \tan\frac{t}{2}),y=a\sin t$ is the catenary $y=a\cosh (\frac{x}{a})$ Prove that the evolute of the tractrix $x=a(\cos t+\log \tan\frac{t}{2}),y=a\sin t$ is the catenary $y=a\cosh (\frac{x}{a})$ * * * Since evolute of a curve is the envelope of the normals of that curve.I first found the normal to the curve. $x=a(\cos t+\log \tan\frac{t}{2}),y=a\sin t$ $\frac{dy}{dx}=\tan t$ So the equation of the normal is $y-a\sin t=-\cot t(x-a\cos t-a\log \tan \frac{t}{2})..........(1)$ Now taking $t$ as a parameter,i tried to find the envelope of the normal by partial differentiating eqn $(1)$ with respect to $t$,we get $-a\cos t=x\csc^2t-a\cot t\sin t-a\cos t\csc^2t+a\frac{\cot t}{\sin t}-a\log \tan \frac{t}{2}\csc^2t...........(2)$ I am stuck here.It has got very complicated.How can i eliminate $t$ between $(1)$ and $(2).$Please help me.

I'm going to use this parametric equation for a tractrix: $x(t)=a(t-\tanh t)$, $y(t)= a \text{ sech } x $

The general formula for an evolute is $$X[x,y]=x-y'\frac{x'^2+y'^2}{x'y''-x''y'}$$ $$Y[x,y]=y+x'\frac{x'^2+y'^2}{x'y''-x''y'}$$

By substituting $x$ with $a(t-\tanh t)$ and $y$ with $a \text{ sech } x$ and simplifying, we get the following: $$X=a*t$$ $$Y=a*\cosh t$$ This is the same as $y=a*\cosh\left(\frac{x}{a}\right)$