Let $$\begin{align*}\gamma:(0,\pi) &\to \mathbb R^2\\ t &\mapsto \gamma(t)=(\sin t,\ \cos t+\log \left(\tan(\frac{t}{2}) \right), \end{align*}$$ be the parametrized curve of the tractrix.

Let $P$ be a point on the tractrix, $L$ the tangent line that passes through $P$, and $Q$ the intersection of $L$ with the $y$ axis. Prove that the distance between $P$ and $Q$ is $1$.

My attempt and questions:

If $P=\gamma(t_0)$, with $t_0 \in (0,\pi)$, then the tangent line $L$ is
$$L: \gamma(t_0)+\lambda\gamma'(t_0).$$

The intersection of the line and the $y$ axists is the point $Q=(0,\gamma_2(t_0)-\dfrac{\gamma_2'(t_0)}{\gamma_1'(t_0)}\gamma_1(t_0)$.

I've tried to find the point of intersection between the tangent line and the $y$ axis and then calculate $\|P-Q \|^2$ but I got stuck.


I've calculated $$\gamma'(t_0)=\left(\cos(t),-\sin(t)+\frac{\sec^2(\frac{t_0}{2})}{2\tan(\frac{t_0}{2})} \right),$$ to know $\gamma_1'(t_0),\gamma_2'(t_0)$ but I couldn't arrive to anything.

Also, what if $\gamma_1'(t_0)=0$?, This can perfectly be the case, for example take $t=\frac{\pi}{2}$. How can I analyze that case separately from the rest?

Any suggestions, ideas, answers would be appreciated.


I'll outline a short proof using only your calculation of $\gamma'(t_0)$. As you will see, it may help to do things one at a time rather than trying to write one big equation and solve for every variable simultaneously. The algebraic details I will leave for you to work out.

Note that $$\frac{\sec^2 \left(\frac{t_0}{2}\right)}{2\tan\left(\frac{t_0}{2}\right)}=\frac{1}{2\sin\left(\frac{t_0}{2}\right)\cos\left(\frac{t_0}{2}\right)}=\csc(t_0)$$ Hence the tangent line (parameterised by $s$) at a point $P=\gamma(t_0)= (x_0,y_0)=\left(\sin(t_0),\cos(t_0)+\log\left(\tan\left(\frac{t_0}{2}\right)\right)\right)$ is given by

$$x=\sin(t_0)+ s\cos(t_0)$$


This line intersects the $y$ axis at $Q=L(s_0)$, where $x=0$. That is,

$$\bigg(\sin(t_0)+t_1(\cos(t_0))=0\bigg)\Longrightarrow s_0=\frac{-\sin(t_0)}{\cos(t_0)}=-\tan(t_0)$$

Substituting this to find $y_1$, the $y$ value of $Q=(0,y_1)$, we get

$$\begin{aligned} y_1 =& \,\cos(t_0)+\log\left(\tan\left(\frac{t_0}{2}\right)\right)-\tan(t_0)\bigg(-\sin(t_0)+\csc(t_0)\bigg)\\ =& \,\log\left(\tan\left(\frac{t_0}{2}\right)\right) \end{aligned}$$

Hence the distance between

$$P=\left(\sin(t_0),\cos(t_0)+\log\left(\tan\left(\frac{t_0}{2}\right)\right)\right)$$ and $$Q=\left(0,\log\left(\tan\left(\frac{t_0}{2}\right)\right)\right)$$ is $\sqrt{\sin^2(t_0)+\cos^2(t_0)}=1$

Note that at $t=\frac{\pi}{2}$, a tangent line doesn't exist, so you can safely ignore that point for the purposes of this question.


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