Step-by-step Solution:

Let F (a,0) be focus and P (\(at^2,2at)\) be point in parabola \[\] 1. Equation: \[ \text{FP} = 5a \] 2. Next equation: \[ a + at^2 = 5a \] Simplifying this: \[ a(1 + t^2) = 5a \] If \(a \neq 0\), we can divide both sides by \(a\): \[ 1 + t^2 = 5 \] Subtracting 1 from both sides: \[ t^2 = 4 \] Taking square roots: \[ t = \pm 2 \] 3. Since \(t > 0\), we have: \[ t = 2 \] 4. Thus, the point \(P\) is given by: \[ P \equiv (4a, 4a) \] 5. Now, we are given a set of options, and the point \(P\) must satisfy one of these options. According to the reasoning, only the option \(P = (1, 1)\) satisfies the condition. \[\] Thus, the correct point is \((1, 1)\).