= Solution
Write $M=m_1+m_2$. Integrating the zero-momentum condition gives $Mx+m_2l\sin\phi=MX_c$, where $X_c$ is the constant horizontal centre-of-mass position. Therefore the bob coordinates are
$$
X=x+l\sin\phi=X_c+\frac{m_1l}{M}\sin\phi,\qquad Y=-l\cos\phi.
$$
Eliminating the angle gives the <elliptic bob path at zero horizontal momentum>,
$$
\boxed{\frac{(X-X_c)^2}{(m_1l/M)^2}+\frac{Y^2}{l^2}=1.}
$$
The semiaxes are $m_1l/M$ horizontally and $l$ vertically. An <oscillation> traverses only the allowed arc, reversing at its <turning points>. The support counter-moves to keep the horizontal <centre of mass> fixed. The heavy-support limit tends to the familiar fixed-pivot circle.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1-sliding-pendulum.png]
{title=Pendulum ellipse and support motion}
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