A tangent ellipse and a bounded area

Very well known equation of ellipse with semi-major and semi-minor axis equal to 5 and 3, respectively, is: $\frac{x^{2}}{25}+\frac{y^{2}}{9} = 1$ This is the equation of ellipse in coordinate system where x and y axis coincide with semi-major and semi-minor axis, respectively, and where center of ellipse coincide with origin of coordinate system. However, when we rotate coordinate axis for some angle $\theta$ and translate the origin to some point $(p,q)$ , the equation of ellipse changes according to these transformations and becomes:

$(x-p)^{2}\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+(y-q)^{2}\left(\frac{\cos^{2}{\theta}}{9}+\frac{\sin^{2}{\theta}}{25}\right)+\frac{16}{225}(x-p)(y-q)\sin{2\theta} = 1$

Differentiating above equation we get equation involving the slope at some point $(x,y)$ on ellipse:

$\begin{aligned} \frac{d}{dx}\, \left[(x-p)^{2}\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+(y-q)^{2}\left(\frac{\cos^{2}{\theta}}{9}+\frac{\sin^{2}{\theta}}{25}\right)-\frac{16}{225}(x-p)(y-q)\sin{2\theta}\right] &= \frac{d}{dx}\, 1\\ 2(x-p)\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+2(y-q)\frac{dy}{dx}\left(\frac{\cos^{2}{\theta}}{9}+\frac{\sin^{2}{\theta}}{25}\right)-\frac{16}{225}\left(y-q+(x-p)\frac{dy}{dx}\right)\sin{2\theta} &= 0\end{aligned}$

We know coordinates of one point - $(2,0)$ - and we know that the slope is 0 at that point. On the other hand, we know that the other point lies on line $y = -x$ which is tangent to ellipse, which provide us with information that the point is of the form $(x,-x)$ and that slope is -1 at that point. Thus, we constitute four equations in for unknowns ( $p,q,\theta,x$ ):

$\begin{aligned} (2-p)^{2}\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+q^{2}\left(\frac{\cos^{2}{\theta}}{9}+\frac{\sin^{2}{\theta}}{25}\right)-\frac{16}{225}(2-p)q\sin{2\theta} &= 1 \\ 2(2-p)\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+\frac{16}{225}q\sin{2\theta} &= 0 \\ (x-p)^{2}\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+(x+q)^{2}\left(\frac{\cos^{2}{\theta}}{9}+\frac{\sin^{2}{\theta}}{25}\right)+\frac{16}{225}(x-p)(x+q)\sin{2\theta} &= 1 \\ 2(x-p)\left(\frac{\cos^{2}{\theta}}{25}+\frac{\sin^{2}{\theta}}{9}\right)+2(x+q)\left(\frac{\cos^{2}{\theta}}{9}+\frac{\sin^{2}{\theta}}{25}\right)-\frac{16}{225}\left(p-q-2x\right)\sin{2\theta} &= 0\end{aligned}$

I've used Matlab's fsolve function with initial guess $(p,q,\theta,x) = (0, 4.5, -\pi/4, -2)$ to solve this non-linear system. Solutions I got were: $(p,q,\theta,x) = (0.080273, 4.163592, -0.806373, -2.013175)$ .

Area is then found to be $A = 0.724965667512301$ and $\left \lfloor 1000A \right \rfloor = 724$ . Proposed correct answer is actually 725 and this slight difference is, I assume, due to accuracy of numerical methods employed.