If y = mx + c is the normal at a point on the parabola y² = 8x whose focal distance is 8 units, then |c| is equal to.

y = mx + c
y² = 8x = 4ax
∴ a = 2
2yy' = 8, ⇒ y' = 4/y
Focal distance of a point (2t², 4t) = 2(1 + t²)
2(1 + t²) = 8
1 + t² = 4
t² = 3
t = ±√3
Slope of tangent at (2(±√3)², 4(±√3)): (6, ±4√3) = 4/y
|y| = ±4√3 = ±√12
Slope of normal = ±√3
m = ±√3
y = ±√3x + c
This line passes through (6, ±4√3)
∴ ±4√3 = ±√3(6) + c
±4√3 = ±6√3 + c
∴ c = ±(4√3 - 6√3) = ±(-2√3)
c = ±10√3
|c| = 10√3