Ellipse described with little information

If angle QRP = $x$ , then $e = \tan (x/2)$ , where e is the eccentricity of the ellipse.

I'll post the complete solution later.

Let the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with focus $S(\sqrt{a^2-b^2},0)$ and latus-rectum endpoints $P(\sqrt{a^2-b^2}, \frac{b^2}{a}), Q(\sqrt{a^2-b^2}, -\frac{b^2}{a})$ . If $\angle{PRQ} = \frac{\pi}{3}$ (with $R$ on the $x-$ axis by symmetry), then the slope of tangent $PR = -\tan(\pi/6) = -\frac{1}{\sqrt{3}}$ and $QR = \tan(\pi/6) = \frac{1}{\sqrt{3}}.$

WLOG, let us focus on tangent $PR$ . If we differentiate the ellipse at $P$ , we obtain:

$\frac{dy}{dx}|_{x=\sqrt{a^2-b^2}} = -\frac{b}{a} \cdot \frac {x}{\sqrt{a^2-x^2}} \Rightarrow -\frac{b}{a} \cdot \frac{\sqrt{a^2-b^2}}{\sqrt{a^2-(a^2-b^2)}} = -\frac{1}{\sqrt{3}}$ ;

or $\frac{a^2-b^2}{a^2} = \frac{1}{3} \Rightarrow b^2 = \frac{2}{3}a^2.$

The ellipse's eccentricity is finally computed per:

$e = \frac{\sqrt{a^2-b^2}}{a} = \frac{\sqrt{(1-2/3)a^2}}{a} = \sqrt{1 - 2/3} = \boxed{\frac{1}{\sqrt{3}}}.$