We have, $\quad S \equiv \frac{x^2}{4}+\frac{y^2}{9}=1$
$$
\therefore \quad A(0, \pm 3) \quad \text { (vertex }=(0, \pm b)
$$

Again,
$$
\begin{aligned}
& S^{\prime} \equiv \frac{x^2}{9}+\frac{y^2}{4}=1 \\
\therefore \quad & F( \pm \sqrt{5}, 0) \quad \text { (Focus }=( \pm a e, 0)
\end{aligned}
$$


Again $P$ is a point on the major axis of the ellipse $S^{\prime}=0$, which divides of in the ratio $2: 1$.
$$
\therefore P=\left(\frac{2 x \pm \sqrt{5}+1 \times 0}{2+1}, \frac{2 \times 0+1 \times 0}{2+1}\right)=\left(\frac{ \pm 2 \sqrt{5}}{3}, 0\right)
$$

Now, equation of line passing through $A$ and $P$ is given by
$$
y-3=\frac{0-3}{\frac{2 \sqrt{5}}{3}-0}(x-0) \Rightarrow y=-\frac{9}{2 \sqrt{5}} x+3
$$

Now, intersection point of line in Eq. (i) and $S=0$ are $(0,3)$ and $\left(\frac{30}{7 \sqrt{5}}, \frac{-6}{7}\right)$.
$\therefore$ Length of chord
$$
\begin{aligned}
& =\sqrt{\left(\frac{30}{7 \sqrt{5}}-0\right)^2+\left(3+\frac{6}{7}\right)^2}=\sqrt{\frac{900}{247}+\frac{729}{49}} \\
& =\sqrt{\frac{4545}{245}}=\sqrt{\frac{909}{49}}=\frac{3 \sqrt{101}}{7} \\
& \therefore \quad k=7 .
\end{aligned}
$$