Question with Solution

With $e=\frac{1}{2}$, the focus (sun) is at distance ae = a/2 from the centre, and $b=a\sqrt{1-e^2}=\frac{\sqrt{3}}{2}a$. Point A is the far vertex, at distance $a+ae=\frac{3a}{2}$ from the sun; let the speed there be $V_2$.

At B (end of the minor axis) the velocity is parallel to the major axis, so its perpendicular distance from the sun is b. Conservation of angular momentum: $m{V}_2\cdot \frac{3a}{2}=m{v}_B\cdot \frac{\sqrt{3}}{2}a$, giving $v_B=\sqrt{3}{V}_2$.

(For the times: equal areal velocity gives $\frac{T_{DAB}}{T_{BCD}}=\frac{\text{Area }FDAB}{\text{Area }FBCD}=\frac{\frac{\pi ab}{2}+\frac{1}{2}\cdot \frac{\sqrt{3}a}{2}\cdot \frac{a}{2}}{\frac{\pi ab}{2}-\frac{1}{2}\cdot \frac{\sqrt{3}a}{2}\cdot \frac{a}{2}}$, neither printed ratio in (A) or (B), so those options are wrong.)

Ans (C)