How do I find the equations of tangents to this Polar Curve? I have this Polar Equation:
$$r = \sin(\theta)+\sqrt{3} \cos(\theta)$$
And I am trying to find the equations of the tangents to this curve that are parallel to the initial line (the line tangent to the curve where $\theta=0$), but I am not sure where to go after this point:
$$y = r\sin(\theta) = \sin(\theta)(\sin(\theta)+\sqrt{3} \cos(\theta))$$
$$\frac{dy}{d\theta} = \sqrt{3} \cos(2\theta)+\sin(2\theta)$$
$$let \frac{dy}{d\theta}=0, \therefore \sqrt{3} \cos(2\theta)+\sin(2\theta) = 0$$
So my questions are these:


*

*Have I done it correctly so far, or have I messed up somewhere?

*If not, how do I solve it from here?

 A: Since you are trying to find the equations of the tangent lines to your curve, your overall approach is wrong. You want to find what are parallel lines in the Cartesian $xy$ plane, so what is relevant is $\dfrac{dy}{dx}$, while $\dfrac{dy}{d\theta}$ is irrelevant. (For a horizontal tangent line you should get $\frac{dy}{d\theta}=0$ but for other lines there is no easy relation.)
One way to solve your problem is to convert the polar equation to Cartesian coordinates and work from there. If you multiply both sides of your beginning equation by $r$ you get terms in $r^2$, $r\cos\theta$, and $r\sin\theta$. You know how to convert those expressions to ones using $x$ and $y$. Do so, and you get the Cartesian equation of a circle.
There are multiple ways to continue from there. One way is to find the Cartesian point where $\theta=0$, use implicit differentiation to find the slope of the tangent line, and use that differentiation to find all other points that get the same slope line.
But there is an easier, more intuitive way. You know that your graph is a circle, and from your earlier studies you should know that the point covers the circle uniformly for $\theta$ between zero and pi. (You may think the terminal point is at $2\pi$, but the values of $r$ become negative after pi so the Cartesian points start repeating at pi.) The parallel lines then happen only at the initial point where $\theta=0$ and at the opposite point $\theta=\pi/2$. The tangent lines there are perpendicular to the line between those two points, which gives you the slope of the tangent lines quickly without calculus. The points and the slope quickly give you the equations of the lines.
