Find the locus of the feet of the perpendiculars dropped

Find the locus of the feet of the perpendiculars dropped from the pole onto the tangents of the logarithmic spiral $\rho=e^{a\varphi}$.

Step 1: Cartesian parametrization and tangent direction

\[ x_0 = e^{a\varphi}\cos\varphi, \quad y_0 = e^{a\varphi}\sin\varphi, \qquad \dot x = e^{a\varphi}(a\cos\varphi-\sin\varphi),\quad \dot y = e^{a\varphi}(a\sin\varphi+\cos\varphi). \tag{1} \]

The tangent's normal direction is $\mathbf n=(\dot y,-\dot x)=e^{a\varphi}(a\sin\varphi+\cos\varphi,\ \sin\varphi-a\cos\varphi)$.

Step 2: Tangent line in normal form

The tangent is $\mathbf n\cdot(x,y)=d$ with $d=\mathbf n\cdot(x_0,y_0)$. Expanding, the bracket $(a\sin\varphi+\cos\varphi)\cos\varphi+(\sin\varphi-a\cos\varphi)\sin\varphi$ collapses to $1$ (the cross terms in $a\sin\varphi\cos\varphi$ cancel), so

\[ d = e^{2a\varphi}, \qquad |\mathbf n|^2 = e^{2a\varphi}(1+a^2). \tag{2} \]

Step 3: The foot $M$

The foot of the perpendicular from the origin to $\mathbf n\cdot\mathbf r=d$ is $M=\dfrac{d}{|\mathbf n|^2}\mathbf n$: \[x_M = \frac{e^{a\varphi}(a\sin\varphi+\cos\varphi)}{1+a^2}, \qquad y_M = \frac{e^{a\varphi}(\sin\varphi-a\cos\varphi)}{1+a^2}. \tag{3}\]

Step 4: Polar radius and angle of $M$

\[ \rho_M^2 = x_M^2+y_M^2 = \frac{e^{2a\varphi}(1+a^2)}{(1+a^2)^2} = \frac{e^{2a\varphi}}{1+a^2} \;\Longrightarrow\; \rho_M = \frac{e^{a\varphi}}{\sqrt{1+a^2}}. \tag{4} \]

\[ \tan\theta_M = \frac{y_M}{x_M} = \frac{\tan\varphi-a}{1+a\tan\varphi} = \tan(\varphi-\arctan a) \;\Longrightarrow\; \varphi = \theta_M+\arctan a. \tag{5} \]

Step 5: Eliminate $\varphi$

\[ \rho_M = \frac{e^{a\arctan a}}{\sqrt{1+a^2}}\cdot e^{a\theta_M} = C\,e^{a\theta_M}, \qquad C=\frac{e^{a\arctan a}}{\sqrt{1+a^2}}. \tag{6} \]

Step 6: Verification

\[a=0.3\text{: for }\varphi=0.4,1.1,2.7,-0.8\text{, the numerically computed foot }M\text{ satisfies }\rho_M=C e^{a\theta_M}\text{ to }\sim10^{-15}\ \checkmark\]