Equation of a trajectory of a point with a constrain on line of sight I am stuck on the following problem:
given a reference frame $x,y$, a particle having speed $|\vec{V}|$ starts from $P_0=(0,y_0)$ and must hit the point $P_T=(x_0,0)$ with the following constrain: the angle between the vector $\vec{V}$ and the line between its position $P(x(t),y(t))$ and $P_T$ must be constant $(\alpha)$ during the flight. What is the equation of the particle trajectory? My attempt is: the cross product between the vector $\vec{V}$ and the vector $\vec{r}$ (vector between $P(x(t),y(t)))$ must be constant. So:
$\vec{V}(t)\cdot\vec{r}=|\vec{V}||\vec{r}|\cos(\alpha)$.
Any suggestion on how to proceed? Thanks
 A: First, to simplify the situation, introduce polar coordinates, centered at the point $P_T \,  (x_0, 0)$, i.e.
\begin{align}
&x = x_0 \, + \, r \, \cos(\theta)\\
&y = r\, \sin(\theta)
\end{align}
The particle in question moves as $\big(x(t), \, y(t)\big)$ so that the velocity
$$\vec{V}(t) \, = \, \left(\frac{dx}{dt}(t), \, \frac{dy}{dt}(t)\right)$$ has constant norm
$$\sqrt{\left(\frac{dx}{dt}(t)\right)^2 \, + \, \left( \frac{dy}{dt}(t)\right)^2} = V$$ and the angle between
$$\vec{V}(t) \, = \, \left(\frac{dx}{dt}(t), \, \frac{dy}{dt}(t)\right) \,\,\text{ and }\,\, \vec{r}(t) \, = \, \big(x(t) - x_0, \, y(t)\big)$$ is constant.
If you switch to polar coordinates, the particle's motion can be written as
\begin{align}
&r(t) = \sqrt{(x(t) - x_0)^2 + y(t)^2}\\
&\theta(t) = \arctan\left(\frac{y(t)}{x(t) - x_0}\right)
\end{align}
Furthermore, recalling the representation of the Riemanian metric in polar coordinates, in these coordinates we get
$$\left(\frac{dx}{dt}\right)^2 \, + \, \left( \frac{dy}{dt}\right)^2 = \left(\frac{dr}{dt}\right)^2 \, + \, r^2 \left(\frac{d\theta}{dt}\right)^2 =  V^2$$ The coordinate lines $\theta = const$ are straight rays with a common origin $P_T \, (x_0, 0)$ while the coordinate lines $r = const$ are concentric circles centered at $P_T \, (x_0, 0)$. At any arbitrary point of the trajectory $(x(t), y(t))$ the angle between the velocity vector $\vec{V}(t)$ and the radial vector $\vec{r}(t)$ is constant $\alpha$. Denote the tangent unit vector at point $(x(t), y(t))$ along the radial vector $\vec{r}$ by $\vec{e}_r = \frac{1}{r}\, \vec{r}$ and the unit vector tangent to the coordinate circle passing through the point $(x(t), y(t))$ by $\vec{e}_{\theta}$. Then the two vectors $\vec{e}_r$ and $\vec{e}_{\theta}$ are orthogonal and unit. Furthermore, by assumption, the angle between $\vec{V}(t)$ and $\vec{e}_r$ is constant $\alpha$, and since the the length of $\vec{V}(t)$ is constant, the complementary angle between $\vec{V}(t)$ and $\vec{e}_{\theta}$ should also be constant $\pi/2 - \alpha$. Consequently, we have the decomposition
$$\vec{V}(t) \, =\, V\cos(\alpha)\, \vec{e}_r \, + \,  V\sin(\alpha)  \, \vec{e}_{\theta}$$ But from the geometric properties of the polar coordinates (e.g. see the Riemannian metric above)
$$\vec{V}(t) \, = \, \frac{dr}{dt} \, \vec{e}_r  \, + \, r\,\frac{d\theta}{dt}  \, \vec{e}_{\theta}$$ which is possible if and only if
\begin{align}
&\frac{dr}{dt} =  V\cos(\alpha)\\
&r\, \frac{d\theta}{dt} =  V\sin(\alpha)
\end{align}
which is the same as the simple system of differential equations
\begin{align}
&\frac{dr}{dt} =  V\cos(\alpha)\\
&\frac{d\theta}{dt} =  \frac{V\sin(\alpha)}{r}\\
& \\
&r(0) = r_0\\
&\theta(0) = \theta_0
\end{align}
where $$r_0 = \sqrt{x_0^2 + y_0^2},\,\,\,\,\,\,\,\theta_0 = \arctan\left(-\,\frac{y_0}{x_0}\right)$$ Then, by solving it, we find
\begin{align}
&r \, = \, r_0\, + \, V\cos(\alpha) \, t\\
&\frac{d\theta}{dt} \, = \,  \frac{V\sin(\alpha)}{r_0\, + \, V\cos(\alpha) \, t}
\end{align}
\begin{align}
r \, &= \, r_0\, + \, V\cos(\alpha) \, t\\
\theta \, &=\, \theta_0 \, - \,\tan(\alpha)\log(r_0) \, + \, \tan(\alpha)\,\log\Big(r_0\, + \, V\cos(\alpha) \, t \Big)\\
\, &= \, \theta_0 \, + \, \tan(\alpha)\,\log\left(\frac{r_0\, + \, V\cos(\alpha) \, t}{r_0} \right)
\end{align}
Finally, the trajectory should be something like this:
\begin{align}
x \, &= \, x_0 \, + \, \Big( r_0\, + \, V\cos(\alpha) \, t\Big)\, \cos\left(\,\theta_0 \, + \, \tan(\alpha)\,\log\left(\frac{r_0\, + \, V\cos(\alpha) \, t}{y_0} \right)\,\right)\\
y \, &= \, \Big( r_0\, + \, V\cos(\alpha) \, t\Big)\, \sin\left(\,\theta_0 \, + \, \tan(\alpha)\,\log\left(\frac{r_0\, + \, V\cos(\alpha) \, t}{r_0} \right)\,\right)
\end{align}
where the time for which the particle reaches $(x_0, 0)$ is $\,t = - \, \frac{r_0}{V\cos(\alpha)}$ so the angle $\alpha$ should be greater than $\pi/2$. Recall
$$r_0 = \sqrt{x_0^2 + y_0^2},\,\,\,\,\,\,\,\theta_0 = \arctan\left(-\,\frac{y_0}{x_0}\right)$$
