Eccentricity Of A Ellipse

From figure we see that phi = 90 degree. So the coordinates of the point P is (ae, b^2/a) So using the formula for tangent at any point (x1,y1) as (xx1)/a^2 +( yy1)/b^2 = 1 We get the slope is -e . From figure -e = tan(150) or e = 0.577 The trick is to just use your concept of polar equation of conic with focus as the pole

We note $\angle PST = \phi = \alpha - \beta = \frac {5\pi}6 - \frac \pi 3 = \frac \pi 2$ . This means that $PS$ is perpendicular to the $x$ -axis at the focus $S$ , implying $P(x_p, y_p) = P(c,y_p)$ , where $c$ is the distance between a focus and the center. Then $y_p$ is given by:

$\begin{aligned} \sqrt{y_p^2 + (2c)^2} + y_p & = 2a \\ \sqrt{y_p^2 + 4c^2} & = 2a - y_p \\ y_p^2 + 4c^2 & = 4a^2 -4a y_p + y_p^2 \\ \implies y_p & = - \frac {a^2-c^2}a = - \frac {b^2}a \end{aligned}$

We note that the tangent at any point on the ellipse is $-\tan (\pi - \alpha) = -\tan \frac \pi 6 = - \frac 1{\sqrt 3}$ . And by differentiating the ellipse equation, we have:

$\begin{aligned} \frac {x^2}{a^2} + \frac {y^2}{b^2} & = 1 & \small \color{#3D99F6} \text{Differentiating w,r.t. }x \\ \frac {2x}{a^2} + \frac {2y}{b^2}\times \frac {dy}{dx} & = 0 \\ \frac {dy}{dx} & = - \frac {b^2x}{a^2y} \\ \implies - \frac {b^2x}{a^2y} & = - \frac 1{\sqrt 3} \\ \frac {b^2{\color{#3D99F6}x_p}}{a^2{\color{#3D99F6}y_p}} & = \frac 1{\sqrt 3} & \small \color{#3D99F6} \text{Note that }x_p = c, \ y_p = \frac {b^2}a \\ \frac {b^2{\color{#3D99F6}ca}}{a^2{\color{#3D99F6}b^2}} & = \frac 1{\sqrt 3} \\ \color{#3D99F6} \frac ca & = \frac 1{\sqrt 3} & \small \color{#3D99F6} \text{Note that }e=\frac ca \\ \color{#3D99F6} e & = \frac 1{\sqrt 3} \\ & \approx \boxed{0.577} \end{aligned}$

Okay, so here, $\alpha =5\pi /6$ , & $\beta =\pi /3$ . Clearly, $\angle PST=\pi -\alpha =\pi -5\pi /6=\pi /6$ . This implies that $\Delta PST$ is a right-angled triangle, with $\phi =\pi /2 (\pi /3+\pi /6=\pi /2$ ; in all right-angled triangles, sum of acute angles is $\pi /2$ .) This is a good thing, as the co-ordinates of the point directly above it can be found easily; $(ae, b^2/a)$ (abscissa of both points are the same, ordinate can be found out).

The tangent's slope is $\tan { 5\pi /6 } =\tan { (\pi -\pi /6) } =-\tan { \pi /6 } =-1/\sqrt { 3 }$ . This slope may be used in the equation for a tangent to an ellipse i.e. $y=mx\pm \sqrt { a^2m^2+b^2 }$ as $m=1/\sqrt { 3 }$ . Once that is done, we can substitute the co-ordinates of the point $(ae, b^2/a)$ to get this scary equation: $\sqrt { 3 } e^{ 2 }+\sqrt { 4-3e^{ 2 } } -e-\sqrt { 3 }$ . I know, it's disturbingly complicated, but it works. You may try to solve it manually, or use WolframAlpha (i used WA, lol). your answer should be 1 or 0.57735. We discard 1 as it is not a valid value of eccentricity for an ellipse. Hence your final answer should be 0.57735, or 0.577 when rounded off. Hope that helped. Corrections are welcomed. :)