A latus rectum of a conic section is a chord through a focus parallel to the directrix. Find the area bounded by the parabola displaystyle{y}={x}^{2}text{/}{left({4}{c}right)} and its latus rectum.

A latus rectum of a conic section is a chord through a focus parallel to the directrix. Find the area bounded by the parabola $y={x}^{2}\text{/}\left(4c\right)$ and its latus rectum.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

FieniChoonin
Step 1
It is known that the area bounded by the curves $y=f\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=g\left(x\right)on\left[a,b\right]$ is given by
$A={\int }_{a}^{b}$ (upper curve - lower curve) dx.
From the figure, the equation of latus rectum is $y=c$ which is upper curve.
Note that the given graph is about y-axis.
Step 2
Substitute $y=c\in y=\frac{{x}^{2}}{4c}$
and obtain that $x=2x$
Thus, the area bounded can be computed as follows.
$A=2{\int }_{0}^{2c}$ (upper curve - lower curve) dx
$=2{\int }_{0}^{2c}\left(c-\frac{{x}^{2}}{4c}\right)dx$
$=2{\left[cx-\frac{{x}^{3}}{12c}\right]}_{0}^{2c}$
$=2\left[2{c}^{2}-\frac{2}{3}{c}^{2}\right]$
$=\frac{8}{3}{c}^{2}$