A planet revolves about the sun in elliptical orbit of semimajor axis $2\times {10}^{12}m.$  The areal velocity of the planet when it is nearest to the sun is $4.4\times {10}^{16}{m}^2/s.$ The least distance between planet and the sun is $1.8\times {10}^{12}m.$ Find the minimum speed of the planet in km/s

Area covered by line joining planet and sun in time dt is $dS=\frac{1}{2}{x}^{2}d\theta$

$\text{Areal velocity = dS /dt =}\frac{1}{2}{x}^{2}d\theta /dt=\frac{1}{2}{x}^2\omega$

where x = distance between planet and sun and $\omega$  = angular speed of planet about sun. From  Keplers second law Areal velocity of planet is constant.

At farthest position $A=dS/dt=\frac{1}{2}{(2R-r)}^2\omega =\frac{1}{2}(2R-r)\left[(2R-r)\omega \right]=\frac{1}{2}(2R-r)V_B$ or 

$V_B=\frac{2A}{2R-r}$ (least speed).  (Using values) $V_B=40km/s.$