The parabola y^2=4x has the normal at the point (4,4). This normal meets the parabola again at a second point. The area of the region enclosed…

At $(4,4)$ the tangent slope is $\dfrac{dy}{dx}=\dfrac{2}{y}=\dfrac12$, so the normal has slope $-2$: $y=-2x+12$. Substituting $x=y^2/4$ gives $y^2+2y-24=0$, so $y=4$ or $y=-6$, i.e. the second intersection is $(9,-6)$. Integrating with respect to $y$, the line lies to the right of the parabola: area $=\displaystyle\int_{-6}^{4}\left(\frac{12-y}{2}-\frac{y^2}{4}\right)dy=\left[6y-\frac{y^2}{4}-\frac{y^3}{12}\right]_{-6}^{4}=\frac{125}{3}$. The value $125/6$ is half of this (a common slip from using a triangle-type formula) and $64/3$ is the area for a different chord. Integrating along the y-axis keeps the region as a single strip, which is simpler than splitting the area at x=4.