= Solution
Using $q=a(1-e)$ and $Q=a(1+e)$, differentiate each apsidal distance. The <apsidal evolution under Poynting–Robertson drag> is
$$
\boxed{\dot q=-\frac{A(1-e)^2(4-e)}{2a(1-e^2)^{3/2}},\qquad
\dot Q=-\frac{A(1+e)^2(4+e)}{2a(1-e^2)^{3/2}}.}
$$
Both the <pericentre distance> and <apocentre distance> decrease. Since $a=(Q+q)/2$ and $e=(Q-q)/(Q+q)$, the same rates are
$$
\dot q=-\frac{A\sqrt q\,(5q+3Q)}{2(Q+q)Q^{3/2}},\qquad
\dot Q=-\frac{A\sqrt Q\,(3q+5Q)}{2(Q+q)q^{3/2}}.
$$
Their ratio is
$$
\boxed{\frac{dQ}{dq}=\left(\frac Qq\right)^2\frac{3q+5Q}{5q+3Q}.}
$$
In particular $Q\gg q$ gives $dQ/dq\simeq(5/3)(Q/q)^2$, whereas the circular limit gives slope $1$.
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