This is an incomplete answer I have only covered the case where an oval is "large" in the sense it contains both of its foci in reasonable details.
To simplify the analysis, let start with $(x_1,y_1) = (-1,0), (x_2, y_2) = (1,0)$ and define
$r, s, u, v$ such that
$$ \begin{cases}r^2 = (x+1)^2 + y^2\\s^2 = (x-1)^2 + y^2\end{cases}
\quad\text{ and }\quad\begin{cases}u = \frac12 (r-s)\\v = \frac12(r+s)\end{cases}
$$
We will look at the area of half of an oval in the upper half plane given by
$$\mathscr{O} = \big\{\; (x,y) \in \mathbb{R}^2 : y \ge 0,\, r + a s \le k \;\big\}$$
WOLOG, we can assume $a \ge 1$. Geometrically, there are three 3 possibilities:
- $k > 2 a$ - The oval is "large" in the sense it encloses both of its foci.
- $k = 2 a$ - The oval turns into a teardrop with a cusp at the focus $(-1,0)$.
- $2a > k > 2$ - The oval encloses one and only one of the focus $(1,0)$.
In the following analysis, we will concentrate on the $1^{st}$ case where $k > 2 a$.
Notice $u^2 + v^2 = \frac12 (r^2 + s^2) = x^2 + 1 + y^2$. On the upper half plane, we have
$$
\begin{cases}
x = \frac14 (r^2 - s^2) = uv\\
\\
y = \sqrt{u^2 + v^2 - u^2 v^2 -1 } = \sqrt{(1-u^2)(v^2-1)}
\end{cases}$$
The area element becomes
$$dx \wedge dy = \frac{dx \wedge d y^2}{2y} = \frac{duv \wedge d(u^2 + v^2)}{2y}
= \frac{(u dv + v du)\wedge(u du + v dv)}{\sqrt{(1-u^2)(v^2-1)}}\\
= \frac{(v^2 - u^2) du \wedge dv}{\sqrt{(1-u^2)(v^2-1)}}
= \frac{( (2v^2-1)+(1-2u^2)) du\wedge dv}{2\sqrt{(1-u^2)(v^2-1)}}
$$
Notice
$$d u\sqrt{1-u^2} = \frac{1-2u^2}{\sqrt{1-u^2}}\quad\text{ and }\quad
d v\sqrt{v^2-1} = \frac{2v^2-1}{\sqrt{v^2-1}}
$$
We get
$$dx \wedge dy = \frac12 d \left[\frac{u(1-u^2)dv - v(v^2-1)du}{\sqrt{(1-u^2)(v^2-1)}}\right]$$
When $k > 2a$, the half oval $\mathscr{O}$ get mapped to a quadrilateral $\mathscr{Q}$ in the $uv$-plane.
$$\mathscr{Q} = \big\{\; (u,v) \in \mathbb{R}^2 : -1 \le u \le 1; 1 \le v \le \alpha + \beta u \;\}$$
where $ \displaystyle \alpha = \frac{k}{a+1},\;\beta = \frac{a-1}{a+1}$.
This gives us
$$2 \text{Area}(\mathscr{O}) = 2\int_{\mathscr{O}} dx \wedge dy
= \int_{\partial\mathscr{Q}} \frac{u(1-u^2)dv - v(v^2-1)du}{\sqrt{(1-u^2)(v^2-1)}}
$$
It is easy to check on $\partial\mathscr{Q}$, the horizontal and the two vertical edges contribute nothing.
As a result, the area of the full oval can be rewritten as
$$2 \text{Area}(\mathscr{O}) = \int_{-1}^1 \frac{(\alpha+\beta u)((\alpha+\beta u)^2-1) - \beta u(1-u^2)}{\sqrt{(1-u^2)((\alpha+\beta u)^2-1)}} du\tag{*1}
$$
Abusing the notation, we will use $v$ as a shorthand for $\alpha + \beta u$ from now on.
Consider the quadratic equation $t = (\alpha+\beta t)(\alpha t + \beta)$. When $k > 2a$, one can verify above equation has two real roots $\eta$ and $\eta^{-1}$ for $t$. We will let $\eta$ be the unique root with $|\eta| < 1$.
Define $\lambda, \mu$ and introduce a new variable $z$ such that
$$\begin{cases}
\lambda &= \alpha + \beta \eta\\
\mu &= \alpha \eta + \beta
\end{cases}
\quad\text{ and }\quad
u = \frac{z + \eta}{1 + \eta z}\;\;\iff\;\;
z = \frac{u - \eta}{1 - \eta u}
$$
One can show that $|\mu| < 1$ and we have
$$
du = \frac{(1-\eta^2)}{(1+\eta z)^2}dz,\quad
\begin{cases}
u & = \frac{z+\eta}{1+\eta z }\\
1 + u & = \frac{(1+\eta)(1+z)}{1+\eta z}\\
1 - u & = \frac{(1-\eta)(1-z)}{1+\eta z}
\end{cases}
\quad\text{ and }\quad
\begin{cases}
v &= \frac{\lambda + \mu z}{1 + \eta z}\\
v + 1 &= \frac{(\lambda+1)(1 + \mu z)}{1+\eta z}\\
v - 1 &= \frac{(\lambda-1)(1 - \mu z)}{1+\eta z}
\end{cases}$$
This implies
$$\begin{align}
v(v^2 - 1) & = \frac{\lambda^2 - 1}{(1+\eta z)^3}(\lambda + \mu z)(1 - \mu^2z^2)\\
u(1-u^2) &= \frac{1 - \eta^2}{(1+\eta z)^3}( z + \eta )(1 - z^2)\\
\frac{du}{\sqrt{(1-u^2)(v^2-1)}} &= \sqrt{\frac{1-\eta^2}{\lambda^2-1}}\frac{dz}{\sqrt{(1-z^2)(1-\mu^2 z^2)}}
\end{align}$$
Notice $\beta( 1 - \eta^2 ) = \mu - \lambda \eta = \mu ( 1 - \lambda^2)$, we can rewrite $(*1)$ as
$$2 \text{Area}(\mathscr{O}) =
\sqrt{(1-\eta^2)(\lambda^2-1)} \int_{-1}^1 \frac{P(z) dz}{
(1+\eta z)^3 \sqrt{(1-z^2)(1-\mu^2 z^2)}}\tag{*2}$$
where $P(z) = (\lambda + \mu z)(1 -\mu^2 z^2) + \mu (z + \eta)(1-z^2)$.
By brute force, one can show that
$$\frac{P(z)}{(1+\eta z)^3} = A(\eta,\mu) + \left[ B(\eta,\mu) + C(\eta,\mu)\frac{\partial}{\partial \eta} + D(\eta,\mu)\frac{\partial^2}{\partial \eta^2}\right]\frac{1}{1+\eta z}$$
where
$$\begin{cases}
A(\eta,\mu) &= -\frac{\mu^3+\mu}{\eta^3},\\
B(\eta,\mu) &= \frac{\mu^4 + (\eta^4+1)\mu^2+\eta^4}{\eta^3\mu},\\
C(\eta,\mu) &= -\frac{{\mu}^{4}+\left( -2\,{\eta}^{4}+2\,{\eta}^{2}+1\right) \,{\mu}^{2}-2\,{\eta}^{4}}{{\eta}^{2}\,\mu}\\
D(\eta,\mu) &= \frac{{\mu}^{4}+\left( {\eta}^{4}-4\,{\eta}^{2}+1\right) \,{\mu}^{2}+{\eta}^{4}}{2\,\eta\,\mu}
\end{cases}$$
Substitute this in $(*2)$, then in terms of following complete elliptic integrals of $1^{st}, 2^{nd}$ and $3^{rd}$ kind:
$$\begin{align}
K(\mu) &= \int_0^1 \frac{dz}{\sqrt{(1-z^2)(1-\mu^2 z^2)}}\\
E(\mu) &= \int_0^1 \sqrt{\frac{1-\mu^2 z^2}{1-z^2}} dz\\
\Pi(\eta^2,\mu) &= \int_0^1 \frac{dz}{(1 - \eta^2 z^2)\sqrt{(1-z^2)(1-\mu^2 z^2)}}
\end{align}$$
We have
$$2\text{Area}(\mathscr{O}) = 2\sqrt{(1-\eta^2)(\lambda^2-1)}
\left(
A K(\mu) + \left[ B + C \frac{\partial}{\partial \eta} + D\frac{\partial^2}{\partial \eta^2}\right]\Pi(\eta^2,\mu)
\right)$$
Please note that the partial derivatives of $\Pi(\eta^2,\mu)$ can be expressed in terms of the complete elliptic integrals itself.
$$\frac{\partial\Pi(n,\mu)}{\partial n} = \frac{1}{2n(\mu^2 - n)(n-1)}\left(n E(\mu) + (\mu^2 - n)K(\mu) + (n^2 - \mu^2)\Pi(n,\mu)\right)$$
So in principle, we can get rid of the partial derivatives above and express the area of the oval solely in terms of complete elliptic integrals directly.
Let us switch to the $3^{rd}$ case where $2 < k < 2a$. The oval now intersect the
line segment joining the two foci and encloses one of the focus $(1,0)$.
Let $\displaystyle \sigma = \frac{1-\alpha}{\beta}$.
On the $uv$-plane, the line $v = \alpha + \beta u$ intersect the line $v = 1$ at
$( \sigma, 1 )$ before the line $u = -1$. The half oval $\mathscr{O}$ get mapped to a triangle $\mathscr{T}$:
$$\mathscr{T} = \big\{\; (u,v) \in \mathbb{R}^2 : u \le 1; 1 \le v \le \alpha+\beta u\;\big\}$$
Once again, we can express the area of half oval as a line integral over $\partial \mathscr{T}$ and the horizontal and vertical edges contributes nothing. As a result,
the area of the full oval is given by
$$2\text{Area}(\mathscr{O}) = \int_\sigma^1
\frac{(\alpha+\beta u)((\alpha + \beta u)^2 - 1) - \beta u (1-u^2)
}{\sqrt{(1-u^2)((\alpha + \beta u)^2-1)}}\tag{*1'}
$$
Since the lower limit $\sigma$ now depends on $\alpha, \beta$. We need a different
Mobius transform which map between $u$ and a new variable $\tilde{z}$ such that
$$\begin{array}{rcr}
u& &\tilde{z}\\
\frac{-1-\alpha}{\beta} & \longleftrightarrow & -\frac{1}{\tilde{\mu}}\\
-1 & \longleftrightarrow & \frac{1}{\tilde{\mu}}\\
\sigma = \frac{1-\alpha}{\beta} & \longleftrightarrow & -1\\
1 & \longleftrightarrow & 1
\end{array}$$
to bring the integrand back to the canonical form. If one perform this substitution,
one will obtain something very similar to $(*2)$ above. The derivation will be very complicated and I will stop here. However, it is sort of clear if one do the dirty work along this direction, ultimately we will be able to express the area of oval again in terms of complete elliptic integrals.
Update
Back to original problem. Notice the distance between $(2,3)$ and $(3,1)$ is $\sqrt{5}$.
If we scale the oval by a factor $\frac{2}{\sqrt{5}}$, it will reduce to the case 3.
we have covered before with following parameters:
$$( k, a ) = (\frac{8}{\sqrt{5}}, 2 )\quad\implies\quad(\alpha,\beta,\sigma) =
( \frac{8}{3\sqrt{5}}, \frac13, 3 -\frac{8}{\sqrt{5}})$$
The area of the original oval is given by the integral in $(*1')$ scaled by a factor
$\left(\frac{\sqrt{5}}{2}\right)^2$:
$$\text{Area} = \frac{5}{4} \int_{3 -\frac{8}{\sqrt{5}}}^1 \frac{\left( {\left( \frac{u}{3}+\frac{8}{3\,\sqrt{5}}\right) }^{2}-1\right) \,\left( \frac{u}{3}+\frac{8}{3\,\sqrt{5}}\right) -\frac{u\,\left( 1-{u}^{2}\right) }{3}}{\sqrt{\left( {\left( \frac{u}{3}+\frac{8}{3\,\sqrt{5}}\right) }^{2}-1\right) \,\left( 1-{u}^{2}\right) }} du$$
Throwing this expression to WA gives us $$3.2040508691540993294644483966405124836692674716427904\cdots$$
