Derivation of a general tractrix

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I would like to derive the differential equation for a general tractrix in parameter form. For me, it is quite obvious that $$\mathbf{A}(t)=\mathbf{P}(t)+\frac{\dot{\mathbf{P}}(t)}{|\dot{\mathbf{P}}(t)|}·d,$$ where $\mathbf{A}(t)$ is the curve of the "puller" and $\mathbf{P}(t)$ is the tractrix and $d$ is the distance between the the "puller" and the "pulled object". On Wikipedia, one can find the closed solution of the tractrix if the puller moves along the $y$ axis, so I first tried to solve that problem and then advance to the more general case.

So $\mathbf{A}(t) = \begin{pmatrix} 0 \\ t \end{pmatrix}$ with the parameter $t$, and $\mathbf{P}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$ and $\mathbf{P}(0) = \begin{pmatrix} d \\ 0 \end{pmatrix}$.

I tried to simply plug in those equations into the general tractrix equation, which yields the following: $$\begin{pmatrix} 0 \\ t \end{pmatrix} = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix} + \frac{d}{\sqrt{\dot{x}^2(t) + \dot{y}^2(t)}} \cdot \begin{pmatrix} \dot{x}(t) \\ \dot{y}(t) \end{pmatrix},\ x(0) = d,\ y(0) = 0.$$

But how shall I proceed from here? Obviously I have found the differential equation for the tractrix, but it looks quite different to those one can find e.g. on Wikipedia, and further, I don't know how to solve the ODE.

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I would say

$x=-\frac{a \dot{x}}{\sqrt{\dot{x}^2+\dot{y}^2}}, \quad y- t = -\frac{a \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}\quad\Rightarrow\quad x^2+(y-t)^2=a^2 \quad \Rightarrow \quad y - t= \pm \sqrt{a^2-x^2}\quad\Rightarrow\quad \pm\sqrt{a^2-x^2}=-\frac{a \dot{y}}{\sqrt{\dot{x}^2+\dot{y}^2}}$

by dividing the last and first equality we will get

$\frac{dy}{dx}=\pm \frac{\sqrt{a^2-x^2}}{x}$

which is a pair of separated differential equations in cartesian coordinates.

Edit - added:

a) $\displaystyle y = \pm \int \frac{\sqrt{a^2-x^2}}{x}\,dx \Rightarrow$ solution in shape $y=\pm f(x)$, for y > 0 and y < 0

b) parametric equations, such as:

$x = a \sin t$ (= appropriate choice)

$\displaystyle \Rightarrow y=\int \frac{\sqrt{a^2-x^2}}{x}\,dx=\cdots=a\int \frac{\cos^2t}{\sin t} \, dt=\cdots$