We will see all points in $\displaystyle (r,\theta)$ format.

Clearly $\displaystyle A = (a(1+\cos \alpha),\alpha)$ and $\displaystyle B = (2a \sec \alpha, \alpha)$.

$\displaystyle OA = a(1+\cos \alpha)$ and $\displaystyle OB = 2a \sec \alpha$.

So $\displaystyle OB = 2OA \Rightarrow 2a \sec \alpha = 2a(1+\cos \alpha) \Rightarrow \cos^2 \alpha + \cos \alpha - 1 = 0$

Solving the quadratic equation, discarding the invalid root, we can get our answer...

**P.S:** Well I am not sure of it but it looks correct, if you have any doubts, just ask ok?It will help me too