Rotate to meet the line

The $x$ -coordinate of the left focus is $-\sqrt{7^2-3^2}=-2\sqrt{10}$ , so the equation of the translated ellipse is $\frac{\left(x-2\sqrt{10} \right)^2}{7^2}+\frac{y^2}{3^2}=1$

Now, instead of rotating the ellipse anticlockwise, let's rotate the line clockwise about the origin through the angle $\theta$ . To do this, we use the map $x\mapsto x\cos(\theta)-y\sin(\theta);\;\;\;y \mapsto x \sin(\theta) + y \cos(\theta)$

so the equation of the rotated line is $x \sin(\theta) + y \cos(\theta) = 16-x \cos(\theta) + y \sin(\theta)$

which rearranges to $y=\frac{x(\cos(\theta)+\sin(\theta))-16}{\sin(\theta)-\cos(\theta)}$

To find the intersection of the line and the ellipse, we substitute this expression for $y$ into the equation of the ellipse: $\frac{\left(x-2\sqrt{10} \right)^2}{7^2}+\frac{\left[ \frac{x(\cos(\theta)+\sin(\theta))-16}{\sin(\theta)-\cos(\theta)} \right]^2}{3^2}=1$

Although this looks pretty hideous, it's just a quadratic in $x$ . Since we want the line to be tangent to the ellipse, we want this quadratic to have exactly one root, so its discriminant must be zero. After some manipulation, this gives the equation $\frac{4}{441} \csc^2 \left(\frac{\pi}{4} - \theta\right) \left(-119 + 64 \sqrt5 \sin\left(\frac{\pi}{4} + \theta\right) \right)=0$

Dividing through, $\begin{aligned} -119 + 64 \sqrt5 \sin\left(\frac{\pi}{4} + \theta\right)&=0 \\ \sin\left(\frac{\pi}{4} + \theta\right) &= \frac{119}{64\sqrt5} \\ \theta &= \sin^{-1} \frac{119}{64\sqrt5} - \frac{\pi}{4} \end{aligned}$

which works out to be around $11.25703^\circ$ giving the answer $\boxed{1125}$ .