AP EAPCET 23-Aug-2021 Shift 2 Paper

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