= Solution
Within this analytic quartic <Landau theory>, equilibrium of a nonconserved scalar <order parameter> requires $f'=0$. At a <continuous phase transition>, its restoring curvature vanishes, so $f''=0$. A nonzero $f'''$ would leave a cubic leading term in the <Taylor expansion> about that state and prevent it from being a local minimum. Thus a stable <critical endpoint of a quartic Landau free energy> requires all three derivatives to vanish, with $f''''>0$. The condition $f''=0$ alone can instead identify a <spinodal point>.
For $f=\bar b\lambda^4+\bar c\lambda^3+\bar a\lambda^2+\bar d\lambda$, $\bar b>0$, the successive equations are
$$
f'''=24\bar b\lambda+6\bar c=0,
\quad f''=12\bar b\lambda^2+6\bar c\lambda+2\bar a=0,
\quad f'=4\bar b\lambda^3+3\bar c\lambda^2+2\bar a\lambda+\bar d=0.
$$
They give
$$
\boxed{\lambda_c=-\frac{\bar c}{4\bar b},
\qquad \bar a_c=\frac{3\bar c^2}{8\bar b},
\qquad \bar d_c=\frac{\bar c^3}{16\bar b^2}}.
$$
At these values, the <free energy> is exactly $f(\lambda_c)+\bar b(\lambda-\lambda_c)^4$, proving stability. With no <electric field>, $\bar d=0$, so necessarily $\bar c=0$, followed by $\lambda_c=\bar a_c=0$.
For $\bar c<0$ and $\chi>0$, the <electric-field coupling to nematic order> supplies the required negative $\bar d_c$ at
$$
\boxed{E_c^2=-\frac{\bar c^3}{16\chi\bar b^2},
\qquad E_c=\frac{(-\bar c)^{3/2}}{4\sqrt\chi\,\bar b}}>0.
$$
Tuning the quadratic coefficient to $\bar a_c$ simultaneously reaches the <field-induced critical endpoint of the isotropic-nematic transition>. The PDF's $\tilde c$ in the last clause is undefined; it is interpreted here as the previously defined $\bar c$. The endpoint is a prediction of the stable quartic, small-field <mean-field approximation>; higher-order terms or a field beyond the expansion's validity can shift it.
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