Given, Ratio of Area of rectangle formed by end points of L.R to that of ellipse is $ \frac{1}{\pi} $. Therefore,
Area of rectangle $A_1$ = Distance between focii × Length of L.R $A_1 = (2ae)(\frac{2b^2}{a}) = 4eb^2$
Area of ellipse $A_2 = \pi ab$, then:- $$ \frac{4eb^2}{\pi ab} = \frac{1}{\pi} $$
Which gives, $ \frac{b}{a} = \frac{1}{4e} $
Now for eccentricity, $$ \frac{b^2}{a^2} = 1 - e^2 $$ $$ \frac{1}{16 e^2} = 1- e^2 $$ Taking $e^2 = t $ and solving, I get,
$$ t = e^2 = \frac{2 \pm \sqrt(3)}{4} $$
I believe I should get that value for $e$ and not $e^2$. Where did I go wrong?
You have not done anything wrong. (D) is the right answer since it squares to $\frac{2\pm\sqrt3}4$: $$\left(\frac{\sqrt3\pm1}{2\sqrt2}\right)^2=\frac{4\pm2\sqrt3}8=\frac{2\pm\sqrt3}4$$