Answer:
Option B
Explanation:
Concept involved
Intersection point of tangents at $(at_1^2,2at_{1})$ and $(at_2^2,2at_{2})$ is ( at1t2 , a/t1 +t2) ) , also tangents drawn at end point of focal chord are perpendicular and intersect on directrix
Since $R\left(-a,a(t-\frac{1}{t})\right)$ lies on y =2x+a
$\Rightarrow$ $ a.(t-\frac{1}{t})=-2a+a$
$\Rightarrow$ $t-\frac{1}{t}=-1$
Thus, length of focal chord
$a( t+\frac{1}{t})^{2}=a\left\{\left(t-\frac{1}{t}\right)^{2}+4\right\}=5a$