= Solution
Write $\mu=\tan\theta=m/k$. The incident down-right ray from $(0,s)$ has slope $-\cot\theta$ and meets the parabolic bottom $H(x)=2x^2$ at $x=x_0\in(0,1/2)$ provided
$$
\boxed{s=2x_0^2+x_0\cot\theta}.
$$
At the impact point the bottom slope is $S=H'(x_0)=4x_0$. A stationary reflection preserves frequency and tangential wavenumber $k+Sm$. For the incident wavevector $(k,m)=a(1,\mu)$ and a reflected up-left ray $(k_r,m_r)=a_r(-1,\mu)$, tangential matching gives
$$
a(1+S\mu)=a_r(S\mu-1).
$$
A positive reflected magnitude $a_r$ therefore exists exactly when $S\mu>1$. This is the supercritical-slope condition for <reflection of an internal-wave ray> and produces a group velocity in the second quadrant.
Since $x_0<1/2$, such a point exists when
$$
\boxed{\arctan(1/2)<\theta\leq\pi/6}.
$$
For any such $\theta$, choose
$$
\frac1{4\tan\theta}<x_0<\frac12,
\qquad
\boxed{\frac{3}{8\tan^2\theta}<s<\frac12+\frac1{2\tan\theta}}.
$$
The initial ray then hits the parabola at a supercritical point and its reflected ray travels up and left, as required.
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