Answer
$area = \frac{8}{3}{c^2}$
Work Step by Step
By Theorem 3, the equation $y = \frac{{{x^2}}}{{4c}}$ is a standard parabola with focus $F=\left(0,c\right)$ and directrix $y=-c$. Since the chord is parallel to the directrix and passes through the focus, the equation of the chord is $y=c$.
We find the limits of integration by finding the intersection of the chord $y=c$ with the parabola. So, we solve the equation:
$y = \frac{{{x^2}}}{{4c}} = c$, ${\ \ \ }$ ${x^2} = 4{c^2}$.
The solution is $x = \pm 2c$.
The area bounded by the parabola and the $x$-axis is
$\mathop \smallint \limits_{ - 2c}^{2c} \frac{{{x^2}}}{{4c}}{\rm{d}}x = \frac{1}{{4c}}\left( {\frac{1}{3}{x^3}} \right)|_{ - 2c}^{2c} = \frac{1}{{12c}}\left( {16{c^3}} \right) = \frac{4}{3}{c^2}$
The area of the region bounded by the vertical lines $x = \pm 2c$, the $x$-axis and the latus rectum is $4c^2$.
So, the area bounded by the parabola and its latus rectum is
$area = 4{c^2} - \frac{4}{3}{c^2} = \frac{8}{3}{c^2}$
