= Solution
Let $h\ge0$ be proportional to the target <probability density function>, with $0<Z=\int h(x)\,dx<\infty$. Define the <ratio-of-uniforms method> through the region
$$
A_h=\{(u,v):u>0,\ u^2\le h(v/u)\}.
$$
Draw $(U,V)$ uniformly with respect to area on $A_h$ and return $X=V/U$. To prove its law, change variables from $(u,x)$ to $(u,v)=(u,ux)$. The <Jacobian determinant> is $u$, and the allowed interval for $u$ is $0<u\le\sqrt{h(x)}$. Consequently
$$
\operatorname{area}(A_h)=\int_{\mathbb R}\int_0^{\sqrt{h(x)}}u\,du\,dx=\frac Z2.
$$
The joint density of $(U,X)$ is therefore $2u/Z$ on that interval. Integrating out $u$ gives
$$
\boxed{f_X(x)=\int_0^{\sqrt{h(x)}}\frac{2u}{Z}\,du=\frac{h(x)}Z.}
$$
For an already normalized density, $Z=1$ and the region has area $1/2$. The name refers to the two coordinates of a uniformly sampled region; drawing two unrelated uniforms on arbitrary intervals and taking their ratio would not give the target law.
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