Let $C$ be the rose defined in polar coordinates as $r = 2\cos(2\theta), \theta \in [0,2\pi]$.
- Find all points $(x(\theta), y(\theta))$ in $C$, for which the slope of the tangent line to $C$ at $(x(\theta),y(\theta))$ is equal to $\tan(\theta)$.
- Deduce from part 1 that there is no circle $S$ with positive radius centered at the origin, for which $C$ and $S$ would intersect perpendicularly at some point.
From what I understand, if there were some $\theta$ such that the slope of $C$ at $\theta$ was equal to $\tan(\theta)$ Then part 2 of this question would be wrong, because $y=x\cdot \tan(\theta)$ is a normal line to any circle at $\theta$. So the only way part 2 is true, is if there are no solutions to part 1, or if the solutions are any $\theta$ such that $r=0$.
However, attempting part 1 we get $$x(\theta) = 2\cos(2\theta)\cos(\theta)$$ $$y(\theta) = 2\cos(2\theta)\sin(\theta)$$ $$\frac{dy}{dx} = \lim_{h\to 0} \frac{y(\theta + h) - y(\theta)}{x(\theta+h) - x(\theta)}$$ which using L'Hôpital's rule to evaluate the limit (by taking the partial derivative with respect to $h$ and then plugging in $h=0$ gives us $$\frac{dy}{dx} = \frac{y'(\theta)}{x'(\theta)}$$ Actually doing the derivative and setting this equal to $\tan(\theta)$ finally gives us the equation $$\tan(\theta) = \frac{4\cos(2\theta)\sin(\theta)-2\cos(2\theta)\cos(\theta)}{2\cos(2\theta)\sin(\theta)+4\sin(2\theta)\cos(\theta)}$$
Playing around with trig identities for a while gives us some answers, plugging it into wolfram alpha gives different answers, probably as I made a mistake, but in either case, we do get 4 solutions, and none of them are ones that make $r=0$. So I don't understand how part 2 could be correct.