The correct option is C 12√3
For paarabola y2=4ax equation of normal at P(at21,2at1) is y=−t1x+2at1+at31 ⋯(1)
here in y2=8x, a=2
now equation (1) will be y=−t1x+4t1+2t31 ⋯(2)
But given normal equation is y=x√2−8√2 ⋯(3)
By comparing (2)&(3)
t1=−√2
So the point will be P(4,−4√2)
as given that normal at P(at21,2at1) passes through Q(at22,2at2) so,
t2=−t1−2t1 ⋯(4)
Now by equation (4), t2=2√2
So point Q≡(16,8√2)
By distance formula we get, PQ=12√3 units