The parabola y^2=4x and its normal at the point (4,4) enclose a bounded region. Its area is

For $y^2=4x$ the point $(4,4)$ corresponds to the parameter $t=2$ in $(t^2,2t)$, and the slope of the tangent there is $\dfrac{dy}{dx}=\dfrac{2}{y}=\dfrac12$. The normal has slope $-2$: $y-4=-2(x-4)$, i.e. $x=6-\dfrac y2$. Intersecting with $x=\dfrac{y^2}{4}$ gives $y^2+2y-24=0$, so $y=4$ or $y=-6$ (the second meeting point is $(9,-6)$). Integrating with respect to $y$, the area is $\displaystyle\int_{-6}^{4}\left(6-\frac y2-\frac{y^2}{4}\right)dy=\left[6y-\frac{y^2}{4}-\frac{y^3}{12}\right]_{-6}^{4}=\frac{125}{3}$. Integrating along $x$ would require splitting the region, which is why $y$ is the better variable.