$$y^2=-b^2\left(1-\frac{x^2}{a^2}\right)$$
$$y^2=\frac{b^2}{a^2}(x^2-a^2)\tag{factor $1/a^2$ out}$$
$$y=\pm\sqrt{\frac{b^2}{a^2}(x^2-a^2)}$$
$$y=\pm\frac{b}{a}\sqrt{x^2-a^2}$$
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".
Best Answer
An asymptote denotes a tendency, when a variable goes to infinity.
When $x$ or $y$ tends to infinity the finite terms contribute only insignificantly . Dividing by $y^2$
$$ \frac{x^2}{a^2 y^2} - \frac{1}{b^2} = \frac{1}{y^2} $$
In order to trace to what curve the graph tends to, we need to a priori set the small finite fraction ... including the increasing denominator ... to zero.