= Solution
For unit <mass>, the <inverse-square force> has <potential energy> $-k/r$, with zero potential at infinity. Conservation of <angular momentum> gives $r^2\dot\theta=h$. Differentiate the polar equation of the <Kepler orbit>:
$$
\frac{dr}{d\theta}=\frac{\ell e\sin\theta}{(1+e\cos\theta)^2},
\qquad
\dot r=\frac{he\sin\theta}{\ell}.
$$
Consequently the total <mechanical energy> is
$$
\begin{aligned}
E&=\frac12\left(\dot r^2+\frac{h^2}{r^2}\right)-\frac{k}{r}\\
&=\frac{h^2}{2\ell^2}
\left[e^2\sin^2\theta+(1+e\cos\theta)^2-2(1+e\cos\theta)\right]\\
&=\boxed{\frac{h^2(e^2-1)}{2\ell^2}},
\end{aligned}
$$
where $k=h^2/\ell$. This also explains the sign distinction between an <elliptic orbit>, a <parabolic Kepler orbit>, and a <hyperbolic Kepler orbit>.
At the original <periapsis>, the <parabolic Kepler orbit> has $e=1$, $r=d=\ell/2$, and purely tangential speed $V$. Thus
$$
\ell=2d,\qquad h=dV,\qquad V^2=\frac{2k}{d}.
$$
The outward radial <impulse> has zero moment about the Sun, so it preserves <angular momentum> and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new <mechanical energy> is $E'=\alpha^2V^2/2$. Applying the energy formula gives
$$
\frac{V^2}{8}(e'^2-1)=\frac{\alpha^2V^2}{2},
\qquad
\boxed{e'=\sqrt{1+4\alpha^2}}.
$$
For $\alpha>0$, the new orbit is a <hyperbolic Kepler orbit>; $\alpha=0$ leaves the parabola unchanged.
To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The <eccentricity vector> is
$$
\mathbf e'=\frac{\mathbf v'\times\mathbf h}{k}-\hat{\mathbf r}
=(1,-2\alpha).
$$
It points towards the new <periapsis>, which is rotated clockwise through $\beta=\arctan(2\alpha)$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit is
$$
\boxed{r=\frac{2d}{1+\cos\theta-2\alpha\sin\theta}
=\frac{2d}{1+e'\cos(\theta+\beta)}}.
$$
In particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-4-comet-orbits.png]
{title=Original parabolic orbit and the orbit after an outward radial impulse}
{description=Illustrative case $\alpha=1/2$, with distances in units of $d$. The solid orange curve is the subsequent hyperbolic trajectory; its dashed continuation is the earlier part of the mathematical conic. The original velocity and radial velocity increment are shown at the impulse point.}
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