JEE Main 16 March 2021 Shift 1 Solved Paper

Given, equation of parabola ⇒y2=2x
Equation of normal of parabola, y2=4ax is tx+y=2at+at3
Here, 4a=2
So, a=1∕2
So, equation of normal ⇒tx+y=t+t3∕2
t3+(2−2x)t−2y=0
As, there are three normals which are passing through (a,0), so there must be three roots of this equation. Let's say t1,t2,t3.
t3+(2−2a)t−2.0=0
⇒‌‌t3+(2−2a)t=0
∴‌‌t1+t2+t3=0
and t1t2+t2t3+t3t1=2−2a
t1,t2,t3∈R
t12+t22+t32>0
(t1+t2+t3)2−2(t1t2+t2t3+t3t1)>0
⇒0−2(2−2a)>0⇒4a>4
⇒a>1