How to solve a simple non-linear optimisation problem in order to find the minimum point in a hyperbola? Consider the equation
$$0.26639x-0.043941y+(5.9313\times10^{-5})xy-(3.9303\times{10^{-6}}) y^2-7242.0404=0$$
with $x,y>0$. If you plot it, it'll look like below:

Now, I want to find a minimum point on this hyperbola, such that $x+y$ is a minimum. 
In other words:
$$\min(x+y)$$
$$Constraints: $$
$$0.26639x-0.043941y+(5.9313\times10^{-5})xy-(3.9303\times{10^{-6}}) y^2-7242.0404=0$$
Any help on how to mathematically find this point would be really helpful.
I've asked a similar question here, but in this one, I wanted to find a corner point such that the hyperbola has the maximum curvature. But this is not the case in this question.
 A: Our goal, as I mentioned in my first comment, is to find the line of the form $y=-x+c$ that minimizes $c,$ and which intersects your hyperbola in the first quadrant.


*

*Find $y'(x).$ I'm actually going to view $x=x(y)$ and find $dx/dy.$ Note that if $dy/dx=-1,$ then $dx/dy=-1$ as well. Moreover, in the other post, I found $x(y)$ explicitly. We have:
\begin{align*}
x(y)&=\frac{7242.0404+\left(3.9303\times{10^{-6}}\right) y^2+0.043941y}{0.26639+\left(5.9313\times10^{-5}\right)\!y} \\
x'(y)&=\frac{0.0662637 y^2+595.215 y-1.18771\times 10^8}{(1. y+4491.26)^2}
\end{align*}

*Set $x'(y)=-1.$ We have 
$x'(y)=-1 \implies y=6122.12, \; x=12165.6.$
This solution is in the first quadrant, so we see that there is a point satisfying our needs. Note: this was also my first approach in the previous problem, but it didn't give quite as good a result for that problem. 

*Now, we only need find the $c$ that serves as the $y$ intersept, and that will be the minimum. We have
\begin{align*}
y&=-x+c \\
6122.12&=-12165.6+c \\
18287.7&=c.
\end{align*}
So that's the minimum, and it occurs at $(12165.6, 6122.12).$
A: Hint.
Assuming the plot gives the restriction shape, the minimum is located in the first quadrant, at the tangency point between the restriction and the line $x+y=\lambda$. Now calling the restriction 
$$
g(x,y) = a x + b y + c x y + d y^2 + e = 0
$$
making the substitution $y = \lambda-x$ we get
$$
a x+b (\lambda -x)+c x (\lambda -x)+d (\lambda -x)^2+e = 0\ \ \ (1)
$$
and after solving for $x$ we get
$$
x = \frac{2 d \lambda\pm \sqrt{(a-b+\lambda  (c-2 d))^2+4 (c-d) (\lambda  (b+d \lambda )+e)}-a+b-c
   \lambda }{2 (d-c)}
$$
but at tangency we have only one solution for $x$ so
$$
(a-b+\lambda  (c-2 d))^2+4 (c-d) (\lambda  (b+d \lambda )+e)=0
$$
for $\lambda = 18287.7$ and after substituting into $(1)$
$$
x = 12165.6
$$
