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
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 $$r_0 = \sqrt{x_0^2 + y_0^2},\,\,\,\,\,\,\,\theta_0 = \arctan\left(-\,\frac{y_0}{x_0}\right)$$