Find the asymptote of the spiral rho²=(a²)/phi, phi>0
Find the asymptote of the spiral $\rho^2=\dfrac{a^2}{\varphi}$, $\varphi>0$.
Explanation of Method
A slower blow-up in ρ drives the height all the way down to zero: the asymptote is the polar axis itself.
Step 1: Cartesian coordinates
With $\rho=|a|/\sqrt\varphi$: \[x=\rho\cos\varphi=\frac{a\cos\varphi}{\sqrt\varphi}, \qquad y=\rho\sin\varphi=\frac{a\sin\varphi}{\sqrt\varphi}. \tag{1}\]
Step 2: Behavior as $\varphi\to0^+$
\[ \lim_{\varphi\to0^+}x=\lim_{\varphi\to0^+}\frac{a\cos\varphi}{\sqrt\varphi}=+\infty. \tag{2} \]
\[ \lim_{\varphi\to0^+}y=a\lim_{\varphi\to0^+}\sqrt\varphi\cdot\frac{\sin\varphi}{\varphi}=a\cdot0\cdot1=0. \tag{3} \]
Step 3: Verification
\[a=1,\ \varphi=0.0001\text{: }x\approx100.0,\ y\approx0.00999998\to0\ \checkmark\]
Result
Final answer:
\[\boxed{\,y=0\,}\]
As $\varphi\to0^+$, $x\to+\infty$ while $y\to0$: the curve approaches the polar axis.