Ellipse's tangent and minimum area

Let the coordinates of $P$ be $(p, q)$ . Since rearranging $\frac{x^2}{2} + y^2 = 1$ gives $y = \frac{1}{2}\sqrt{4 - 2x^2}$ , the coordinates of $P$ are $\Big(p, \frac{1}{2}\sqrt{4 - 2p^2}\Big)$ .

By implicit differentiation on $\frac{x^2}{2} + y^2 = 1$ , $x + 2y\frac{dy}{dx} = 0$ , or $\frac{dy}{dx} = -\frac{x}{2y}$ . The slope of the tangent line at $P$ is then $k = -\frac{p}{2q} = -\frac{p}{\sqrt{4 - 2p^2}}$ , and its equation is $y - \frac{1}{2}\sqrt{4 - 2x^2} = -\frac{p}{\sqrt{4 - 2p^2}}(x - p)$ , or $y = \frac{-px + 2}{\sqrt{4 - 2p^2}}$ .

$Q$ is the intersection of $x = 2$ and $y = \frac{-px + 2}{\sqrt{4 - 2p^2}}$ , so its coordinates are $\Big(2, \frac{2(1 - p)}{\sqrt{4 - 2p^2}}\Big)$ .

The slope of $FP$ is $m_1 = \frac{\frac{1}{2}\sqrt{4 - 2p^2} - 0}{p - 1} = \frac{\sqrt{4 - 2p^2}}{2(p - 1)}$ , and the slope of $FQ$ is $m_2 = \frac{\frac{2(1 - p)}{\sqrt{4 - 2p^2}} - 0}{2 - 1} = \frac{2(1 - p)}{\sqrt{4 - 2p^2}}$ .

Since $m_1 \cdot m_2 = \frac{\sqrt{4 - 2p^2}}{2(p - 1)} \cdot \frac{2(1 - p)}{\sqrt{4 - 2p^2}} = -1$ , $\angle PFQ$ is always a right angle.

The distance $FP = \sqrt{(p - 1)^2 + (\frac{1}{2}\sqrt{4 - 2p^2} - 0)^2} = \frac{p - 2}{\sqrt{2}}$ and the distance $FQ = \sqrt{(2 - 1)^2 + (\frac{2(1 - p)}{\sqrt{4 - 2p^2}} - 0)^2} = \frac{p - 2}{\sqrt{2 - p^2}}$ .

The area of right triangle $\triangle PFQ$ is then $A = \frac{1}{2} \cdot \frac{p - 2}{\sqrt{2}} \cdot \frac{p - 2}{\sqrt{2 - p^2}} = \frac{(p - 2)^2}{2\sqrt{4 - 2p^2}}$ .

Since $\frac{d^2A}{dp^2} > 0$ , the minimum area occurs when $\frac{dA}{dp} = \frac{(p - 2)(p^2 +2p - 4)}{2(p^2 - 2)\sqrt{4 - 2p^2}} = 0$ , which solves to $p = \sqrt{5} - 1$ for $-\sqrt{2} < p < \sqrt{2}$ .

Since $k = -\frac{p}{\sqrt{4 - 2p^2}}$ and $p = \sqrt{5} - 1$ , $k^2 = \frac{1 + \sqrt{5}}{2} = \phi$ and $\lfloor 10000k^2 \rfloor \approx \boxed{16180}$ .