Following Calvin Khor's line of reasoning, we will use the following algorithm:
- Find the center of the hyperbola, and translate the hyperbola so that the center coincides with the origin.
- Find the angle of rotation necessary to put the hyperbola into the canonical form $x^2/a^2-y^2/b^2=1.$
- The corner points are represented, at this point, by $x=\pm a.$
- Rotate these two points back through the angle found in Step 2.
- Translate these two points back through the translation performed in Step 1.
The wikipedia page on the general hyperbola will be our guide, here.
Step 1. According to the wiki page, we must write the hyperbola in the form
$$A_{xx}x^2+2A_{xy}xy+A_{yy}y^2+2B_xx+2B_yy+C=0. $$
We have
$$0x^2+\left(5.9313\times 10^{-5}\right)xy-\left(3.9303\times 10^{-6}\right)y^2 + 0.26639x-0.043941y-7242.0404=0, $$
making
\begin{align*}
A_{xx}&=0 \\
A_{xy}&=\left(5.9313\times 10^{-5}\right)/2=2.96565\times 10^{-5} \\
A_{yy}&=-3.9303\times 10^{-6} \\
B_x&=0.26639/2=0.133195 \\
B_y&=-0.043941/2=-0.0219705 \\
C&= -7242.0404.
\end{align*}
We check for the hyperbola nature, namely, that
$$D=\left|\begin{matrix}A_{xx}&A_{xy}\\ A_{xy} &A_{yy} \end{matrix}\right|<0,\quad\text{or}\quad D=\left|\begin{matrix}0&2.96565\times 10^{-5}\\ 2.96565\times 10^{-5} &-3.9303\times 10^{-6} \end{matrix}\right|=-8.79508\times 10^{-10}<0, $$
which is clearly true. The center $(x_c,y_c)$ of the hyperbola is given by
\begin{align*}
x_c&=-\frac{1}{D}\left|\begin{matrix}B_x &A_{xy}\\ B_y &A_{yy}\end{matrix}\right| =\frac{1}{8.79508\times 10^{-10}}\left|\begin{matrix}0.133195 &2.96565\times 10^{-5}\\ -0.0219705 &-3.9303\times 10^{-6}\end{matrix}\right|=145.618\\
y_c&=-\frac{1}{D}\left|\begin{matrix}A_{xx} &B_x\\ A_{xy} &B_y\end{matrix}\right|
=\frac{1}{8.79508\times 10^{-10}}\left|\begin{matrix}0 &0.133195\\ 2.96565\times 10^{-5} &-0.0219705\end{matrix}\right|=-4491.26.\end{align*}
Step 2. The angle of rotation is given by
\begin{align*}
\tan(2\varphi)&=\frac{2A_{xy}}{A_{xx}-A_{yy}} \\
\varphi&=\frac12\,\arctan\left(\frac{2A_{xy}}{A_{xx}-A_{yy}}\right)=0.752315\,\text{rad}=43.1045^{\circ},
\end{align*}
which definitely looks about right.
Step 3. The formula for $a^2$ is
$$a^2=-\frac{\Delta}{\lambda_1 D}, $$
where
\begin{align*}
\Delta&=\left|\begin{matrix}A_{xx}&A_{xy}&B_x\\A_{xy}&A_{yy}&B_y\\B_x&B_y&C\end{matrix}\right|=6.26559\times 10^{-6} \\
0&=\lambda^2-(A_{xx}+A_{yy})\lambda+D.
\end{align*}
Unfortunately, the wiki page fails to distinguish between $\lambda_1$ and $\lambda_2$. If we examine the signs, we must have $a^2>0,$ which means, since $D<0$ and $\Delta>0,$ we must pick the positive root. We have
\begin{align*}
\lambda_2&=-3.16867\times 10^{-5}\\
\lambda_1&=2.77564\times 10^{-5},
\end{align*}
so that
$$a=\pm 16020.6. $$
Step 4. The point we need to rotate is $(16020.6, 0)$ counter-clockwise through an angle $\varphi=0.752315\,\text{rad}$. The rotation matrix for doing that is given by
$$R=\left[\begin{matrix}\cos(\varphi)&-\sin(\varphi)\\ \sin(\varphi) &\cos(\varphi)\end{matrix}\right]=\left[\begin{matrix}0.730109&-0.683331\\ 0.683331 &0.730109\end{matrix}\right]. $$
After rotating, the point is located at $(11696.8, 10947.4).$
Step 5. This is the moment of truth! We must translate back to the original coordinate system. The center of the original hyperbola was located at $(145.618, -4491.26).$ What we must do is add the coordinates together to get the un-translated version. The final point is located at $(11842.4, 6456.14).$
This is not too far off from my other answer! We check to make sure this point is on the curve
$$x=\frac{7242.0404+\left(3.9303\times{10^{-6}}\right) y^2+0.043941y}{0.26639+\left(5.9313\times10^{-5}\right)\!y}, $$
and it is. So I say that this point is the "corner" of your graph.
