Find the polar coordinates of the points on $r^2 = a^2 \sin 2\theta$ where the tangent is perpendicular to the initial line ($\theta=0).$
I guess they are assuming $a>0.$
The answer in the back of the book is $\left( a\sqrt{\frac{\sqrt{3}}{{2}}}, \frac{\pi}{6} \right),\ \left( a\sqrt{\frac{\sqrt{3}}{{2}}}, \frac{7\pi}{6} \right),\ \left( 0, \frac{\pi}{2} \right).$
The method they want us to use is to set $\frac{dx}{d\theta}=0$ and see what happens.
I think I can deduce the first two polar coordinates, but I don't how to logically deduce that $\left( 0, \frac{\pi}{2} \right)$ is perpendicular to the initial line because all methods I think of involves dividing by $0.$
Edit: For example, the method suggested by the book says we should do the following:
$x=r\cos\theta = a\cos\theta\sqrt{\sin (2\theta)}.$ Now set $\ \frac{dx}{d\theta} = 0.$ This implies $ \frac{ a\cos(2\theta)\cos\theta }{\sqrt{\sin(2\theta)}} - a\sin\theta\sqrt{\sin(2\theta)} = 0.$
But here we see the problem: this cannot be evaluated at $\theta=\frac{\pi}{2},$ which is one of the answers, because it would be dividing by $0$ which is not allowed.
To elaborate, if you tell me that $\frac{dx}{d\theta} = 0$ when $\theta=\frac{\pi}{2},$ my response is, "no, it is $ \frac{ a\cos\left(2\frac{\pi}{2}\right)\cos\frac{\pi}{2} }{\sqrt{\sin\left(2\frac{\pi}{2}\right)}} - a\sin\frac{\pi}{2}\sqrt{\sin\left(2\frac{\pi}{2}\right)}$ , which is undefined."
I figured out the rough shape of the graph by hand by doing a small table of $(r,\theta)$ pairs, although from doing this it still isn't clear what the derivative(s) is (are) at $r=0.$
Here is an image of the graph when $a=5:$
Note that this is an A Level question where polar coordinates have only just been introduced in this chapter, so overly-advanced solutions are not appropriate.
$x=a\sqrt{\sin2\theta}\cos\theta$, $y=a\sqrt{\sin2\theta}\sin\theta$.
$\frac{dx}{dy}=\frac{\frac{dx}{d\theta}}{\frac{dy}{d\theta}}=-\frac{\cos3\theta}{\sin\theta}=0\implies\theta=\frac\pi 6,\frac\pi 2,\frac{7\pi}6.$