= Solution
Use an <affine parameter> $\lambda$, with dots denoting its derivatives, and take $\alpha\ne0$ so the metric is nondegenerate for $r>0$. The <geodesic Lagrangian> is $\mathcal L=(\dot r^2+\alpha^2r^4\dot\phi^2)/2$. The <Euler-Lagrange equations> give
$$
\boxed{\ddot r-2\alpha^2r^3\dot\phi^2=0,\qquad \frac{d}{d\lambda}(\alpha^2r^4\dot\phi)=0,}
$$
or $\ddot\phi+4\dot r\dot\phi/r=0$. The conserved cyclic <angular momentum> is
$$
\boxed{J=\frac{\partial\mathcal L}{\partial\dot\phi}=\alpha^2r^4\dot\phi.}
$$
This is the <cyclic angular momentum in a rotational metric>, associated with the rotational <Killing vector> $\partial_\phi$. Explicitly, $\dot J=\alpha^2(4r^3\dot r\dot\phi+r^4\ddot\phi)=0$ by the angular equation.
Differentiating the <Lagrangian> and substituting both equations gives
$$
\frac{d\mathcal L}{d\lambda}=\dot r\ddot r+2\alpha^2r^3\dot r\dot\phi^2+\alpha^2r^4\dot\phi\ddot\phi=(2+2-4)\alpha^2r^3\dot r\dot\phi^2=0.
$$
Thus \b[the squared tangent norm $2\mathcal L$ is constant]. If $\lambda$ is <arc length>, it equals one; its numerical value otherwise depends on affine normalization. An arbitrary nonaffine parametrization would not preserve this constant-speed statement.
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