The normal to the parabola y^2=4x at P(4,4) meets the parabola again at Q. Let F be the focus of the parabola.

$P(4,4)$ corresponds to the parameter $t_1=2$ on $(t^2,2t)$. The normal at $t_1$ meets the parabola again at $t_2=-t_1-\frac2{t_1}=-3$, so $Q=(9,-6)$; option (A) with $(9,6)$ is wrong. $PQ=\sqrt{5^2+10^2}=5\sqrt5\neq10$, so (B) is wrong. With $F=(1,0)$ the area of $FPQ$ is $\frac12\left|(4-1)(-6-0)-(9-1)(4-0)\right|=\frac12|-18-32|=25$, not $\frac{25}2$, so (C) is wrong. For (D) integrate in $y$ from $-6$ to $4$: the chord is $x=4-\frac12(y-4)$ and the parabola is $x=\frac{y^2}4$; since the difference is a quadratic with leading coefficient $\frac14$, the enclosed area is $\frac{(\Delta y)^3}{6\cdot4}=\frac{1000}{24}=\frac{125}3$: correct. Answer: D only.