Let P be the point (8,8) on the parabola y^2=8x, and let the focal chord through P meet the parabola again at Q.

Write $P=(2t^{2},4t)$ with $t=2$. The tangent $ty=x+2t^{2}$ is $2y=x+8$, meeting the $x$-axis at $(-8,0)$; the normal $y=-tx+4t+2t^{3}$ is $y=-2x+24$, meeting the axis at $(12,0)$. The triangle has base $20$ and height $8$, so area $=80\neq40$ (A false). The other end of the focal chord has parameter $-\frac1t=-\frac12$, i.e. $Q=\left(\frac12,-2\right)$, and the focal-chord length is $x_P+x_Q+2a=8+\frac12+4=\frac{25}{2}$ (B true). The normal at parameter $t$ meets the curve again at $t'=-t-\frac2t=-3$, i.e. at $(18,-12)$, not $(32,-16)$ (C false). A circle on any focal chord as diameter touches the directrix $x=-2$: centre $x=\frac{17}{4}$, radius $\frac{25}{4}$, so the leftmost point is $x=-2$, not $-1$ (D false). Answer: B.