Denote our asymptotes here as $L_{+}: x - y = 0$ (the diagonal line) and $L_{-}: x + y = 0$ the off-diagonal line. We have
\begin{align}
x^2 - y^2 &= a^2 \\
\implies \frac{x + y}{\sqrt{2}} \frac{x - y}{\sqrt{2}} &= \frac{ a^2 }2 \\
\implies D(P, L_{-}) \cdot D(P, L_{+}) &= \frac{ a^2 }2 \tag*{Eq.(1)}
\end{align}
where per the setting $P$ is a point on the hyperbola, and $D(\text{point, line})$ is the perpendicular distance between the point and the line.
This is one of the basic properties of any hyperbola that the product of the distances to the asymptotes are constant.
Denote the midpoint of $PN$ as $M$, then by definition $|MN| = \frac12 |PN|$.
Consider the situation where $N$ is on $L_{+}$, which means
$$D(M, L_{+})=|MN|= \frac12 |PN| = \frac12 D(P, L_{+})\tag*{Eq.(2)} $$
Now, for our rectangular hyperbola, the distance to the other asymptote remain the same
$$D(M, L_{-}) = D(P, L_{-}) \tag*{Eq.(3)}$$
because the asymptotes are perpendicular $L_{+} \perp L_{-}$.
Denote the coordinates of $M$ as $(x_1, y_1)$ and use them in the last step of rewriting Eq.(1):
\begin{align}
\text{take Eq.(2) into Eq.(1)}& & \implies && D(P, L_{-}) \cdot 2 D(M, L_{-}) &= \frac{ a^2 }2 \\
\text{from Eq.(3)}& & \implies && D(M, L_{-}) \cdot D(M, L_{-}) &= \frac{ a^2 }4 \\
&& \implies && \frac{ x_1 + y_1}{ \sqrt{2} } \cdot \frac{ x_1 - y_1}{ \sqrt{2} } &= \frac{ a^2 }4 \\
\end{align}
This gives us the hyperbola $\displaystyle (x_1)^2 - (y_1)^2 = \frac{a^2}2$ with the same asymptotes just with a different "constant".
To justify the approach, note that asymptotically the quadratic terms of the curve
$f(x,y)=3x^2+2xy-y^2+8x+10y+14$ dominate, i.e.
$$3x^2+2xy-y^2=(3x-y)(x+y)=0$$
which corresponds to the asymptotic behaviors of the asymptotes and yields their slopes $3,\> -1$.
To obtain the actual equations of the two asymptotes, let $f’_x= f’_y=0$ to determine the center, i.e.
$$ 6x+2y+8=0,\>\>\>\>\>2x-2y+10=0$$
Solve to get the center $(-\frac94, \frac{11}4)$. Then, use the point-slope formula for the equations
$$y-\frac{11}4=-(x+\frac94),\>\>\>\>\> y-\frac{11}4=3(x+\frac94)$$
Best Answer
Your intuition can be stated as the proposition below:
Proof: Note that this proposition is irrelevant to coordinate systems, thus without loss of generality, the hyperbola can be assumed to be $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
Suppose the coordinates of $A$ and $B$ are $(x_1, y_1)$ and $(x_2, y_2)$, respectively, then $x_1, x_2 ≠ 0$ and$$ \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1, \quad \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1. \tag{1} $$ Because the asymptotes are $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$ and $\left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)$ lies on the asymptotes, then$$ \frac{(x_1 + x_2)^2}{a^2} - \frac{(y_1 + y_2)^2}{b^2} = 0. \tag{2} $$
Now, suppose $x_1 + x_2 ≠ 0$, then $y_1 + y_2 ≠ 0$ by (2). From (1) there is$$ \frac{x_1^2 - x_2^2}{a^2} = \frac{y_1^2 - y_2^2}{b^2}, $$ and from (2) there is$$ \frac{(x_1 + x_2)^2}{a^2} = \frac{(y_1 + y_2)^2}{b^2}, $$ thus\begin{align*} &\mathrel{\phantom{\Longrightarrow}}{} \frac{x_1 - x_2}{x_1 + x_2} = \frac{\dfrac{x_1^2 - x_2^2}{a^2}}{\dfrac{(x_1 + x_2)^2}{a^2}} = \frac{\dfrac{y_1^2 - y_2^2}{b^2}}{\dfrac{(y_1 + y_2)^2}{b^2}} = \frac{y_1 - y_2}{y_1 + y_2}\\ &\Longrightarrow \frac{2x_1}{x_1 + x_2} = \frac{x_1 - x_2}{x_1 + x_2} + 1 = \frac{y_1 - y_2}{y_1 + y_2} + 1 = \frac{2y_1}{y_1 + y_2}\\ &\Longrightarrow \frac{y_1}{x_1} = \frac{y_1 + y_2}{x_1 + x_2} \Longrightarrow \frac{y_1}{x_1} = \frac{y_2}{x_2} := c. \end{align*} Note that $x_1 + x_2 ≠ 0$, plugging $y_1 = cx_1$ and $y_2 = cx_2$ into (2) to get $\dfrac{1}{a^2} - \dfrac{c^2}{b^2} = 0$, then by (1) there is$$ 1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = \left( \frac{1}{a^2} - \frac{c^2}{b^2} \right) x_1^2 = 0, $$ a contradiction. Therefore, $x_1 + x_2 = 0$, which implies $y_1 + y_2 = 0$ by (2). Hence the midpoint of $AB$ is the center $(0, 0)$ of the hyperbola.