The curve $r = ae^{\theta \cot \alpha}$ cuts any radius vector in consecutive points $P_1, P_2,...,P_n, P_{n+1},...$.If $\rho_n$ denotes the radius of curvature at $P_n$, prove that $\frac{1}{m-n}\log\bigg(\frac{\rho_m}{\rho_n}\bigg)$ is a constant for all integral values of $m, n$.
Now, my attempt goes like this, I know this is the polar form of a curve, and so we can write the radius of curvature at any $\theta$ to be
$$\rho = \frac{[r_1^2+r^2]^{3/2}}{r^2+2r_1^2-rr_2}$$
where $r_1 = \frac{dr}{d\theta}$, $r_2 = \frac{d^2r}{d\theta^2}$
But then, what are the expressions for $\rho_m, \rho_n$? The above relation would give me the radius of curvature at any $\theta$ in terms of $a, \alpha, \theta$, but how do I proceed to prove the given expression?
First step, you can easily show that $\rho=cr$, where $c$ is a constant (it just depends on $\alpha$). Then what you need to do is to calculate the intersections of the curve with a given radius vector. What's a radius vector? It's a straight line that has the origin at one end, and goes to infinity in the other direction. You can describe this as all the $(x,y)$ pairs such that $$\theta =\mathrm{atan2}(y,x)$$ The intersections of the original curve with this radius vector are at $$r_n=ae^{(\theta+(n-1)2\pi)\cot\alpha}$$ Can you take it from here?