Finding the equation of a tangent line in polar coordinates?

[Eclair_de_XII]

Homework Statement

"Slopes of tangent lines Find the slope of the line tangent to the following polar curves at the given points. At the points where the curve intersects the origin (when this occurs), find the equation of the tangent line in polar coordinates."

$7.$ $r=8sinθ;(4,\frac{5π}{6})$

Homework Equations

$\frac{dy}{dx}=\frac{f'(θ_0)sinθ_0+f(θ_0)cosθ_0}{f'θ_0)cosθ_0-f(θ_0)sinθ_0}$
Answers: $-\sqrt{3};θ=0$

The Attempt at a Solution

$r=f(θ)=8sinθ$
$\frac{dy}{dx}=\frac{8cosθsinθ+8sinθcosθ}{8cos^2θ-8sin^2θ}=\frac{16cosθsinθ}{8(cos^2θ-sin^2θ)}=\frac{8sin2θ}{8cos2θ}=tan2θ$
$f'(\frac{5π}{6})=tan(\frac{5π}{3})=-\sqrt{3}$

So I got the first part down; I just need to know what they're asking when they say "polar equation of tangent line at origin." I could always graph this, but how would you go about solving for this polar equation of the tangent line without doing so? It doesn't really have a radius, since it has an arbitrarily long length, right? But it has an angle at which it touches $(4,\frac{5π}{6})$. It's asking for when it touches the pole of the graph, though.

 

[Eclair_de_XII]

Wait, I think I just figured this out. When it intersects the origin, $r=0$. So all I had to do was just set the initial equation $r=8sinθ$ to $0=8sinθ$ and the angle of the tangent line at $r=0$ is 0 radians. $r$ denotes the distance of the function from the pole, I think. Could someone check these statements for me? Thanks.