Given parabola, y2=16x ...(i) Comparing with y2=4px
∴ Focus S=(4,0) ∵P(a,2a) is in interior region of the parabola y2−16x=0 ∴(2a)2−16a<0 ⇒a2−4a<0 Also (a,2a) and vertex (0,0) are on the same side of x=4 ⇒x−4=0 So, a−4<0 ⇒a<4 Now, a2−4a<0 ⇒0<a<4