Problem 227

Solution:
$(1)$ According to the problem, the semi-focal distance $c= \sqrt{13}$. Let the semi-major axis of the ellipse be $a$, then the real semi-axis of the hyperbola is $a-4$,
The ratio of eccentricities is $\dfrac{3}{7} = \dfrac{\dfrac{\sqrt{13}}{a}}{\dfrac{\sqrt{13}}{a-4}}$, solving this gives $a=7$,
$\therefore$ The length of the semi-minor axis of the ellipse is $\sqrt{49-13}=6$,
The length of the imaginary semi-axis of the hyperbola is $\sqrt{13-9}=2$,
$\therefore$ The equations of the ellipse and the hyperbola are respectively: $\dfrac{x^2}{49} + \dfrac{y^2}{36} = 1$ and $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$;
$(2)$ According to the definition of an ellipse: $PF_1 + PF_2 = 2a = 14$,
According to the definition of a hyperbola: $PF_1 - PF_2 = 6$,
$\therefore PF_1 = 10$, $PF_2 = 4$,
Also, $F_1F_2 = 2\sqrt{13}$, in $\triangle F_1PF_2$, using the cosine theorem, we get: $(2\sqrt{13})^2 = 100 + 16 - 80\cos \angle F_1PF_2$,
$\therefore \cos \angle F_1PF_2 = \dfrac{4}{5}$, then $\sin \angle F_1PF_2 = \dfrac{3}{5}$.
$\therefore S_{\triangle F_1PF_2} = \dfrac{1}{2}PF_1 \cdot PF_2 \cdot \sin \angle F_1PF_2 = \dfrac{1}{2} \times 10 \times 4 \times \dfrac{3}{5} = 12$.
Thus, the final answers are:
$(1)$ The equations of the ellipse and the hyperbola are $\boxed{\dfrac{x^2}{49} + \dfrac{y^2}{36} = 1}$ and $\boxed{\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1}$ respectively;
$(2)$ The area of $\triangle F_1PF_2$ is $\boxed{12}$.