Show that the trajectory is a straight line.

Question

I'm having difficulty answering an exercise for a course in Vector Calculus.

Consider the following trajectory

$$\mathbf{r}(t) = \frac{A}{\cos(\theta(t)) - B\sin(\theta(t))}\hat{\rho}(\theta(t))$$

where $A$ and $B$ are constants and $\hat{\rho}(\theta(t)) = \cos(\theta(t))\hat{\mathbf{i}} +\sin(\theta(t))\hat{\mathbf{j}}$, and $\theta(t)$ is a general function dependent on time. Show that $\mathbf{r}(t)$ is a straight line.

I've attempted to show it by letting $\cos(\theta(t)) = \frac{x}{r}$ and $\sin(\theta(t)) = \frac{y}{r}$ and some cancellation occurs but I end up with $x - By$ in the denominator, which isn't good.

Answer 1

$$\mathbf{r} = \frac{A\cos \theta}{\cos \theta - B\sin \theta}\hat{\mathbf i} + \frac{A\sin \theta}{\cos \theta - B\sin \theta}\hat{\mathbf j} = (x,y)$$

$$\frac1A x - \frac{B}{A} y = 1 \implies y = \frac1B x - \frac{A}{B}$$

Which shows that $\mathbf{r}$ is a straight line.

Answer 2

If $\vec r=\hat \rho\,\frac{A}{\cos(\theta)-B\sin(\theta)}$, then we have

$$x=\frac{A\cos(\theta)}{\cos(\theta)-B\sin(\theta)}=\frac{Ax}{x-By}\\\\ y=\frac{A\sin(\theta)}{\cos(\theta)-B\sin(\theta)}=\frac{Ay}{x-By}$$

from which we see that

$$x=By+A$$

And we are done!