Question 1110815
{{{y^2=9x}}}
In this case, the latus rectum is part of the line {{{x=9/4=2.25}}} .
Parabola, focus, directrix and latus rectum look like this
{{{drawing(400,400,-3,6,-5,5,
graph(400,400,-3,6,-5,5,3sqrt(x),-3sqrt(x)),
line(-9/4,-6,-9/4,6),blue(circle(9/4,0,0.1)),
blue(line(9/4,9/2,9/4,-9/2)),locate(4.3,2.3,blue(focus)),
locate(-2.15,4,directrix),blue(arrow(4.25,2,2.33,0.08)),
locate(2.35,-2,blue(latus)),locate(2.35,-2.6,blue(rectum))
)}}} The area we want is the area between the red/green curve and the blue latus rectum.
 
ONE WAY:
As the halves above and below the x-axis are symmetrical, we can calculate that area as twice the area above the x-axis.
When we only consider that half, with {{{y>=0}}} ,
{{{y^2=9x}}} is equivalent to {{{y=3sqrt(x)=3x^"1 / 2"}}} , so {{{area=2}}}{{{int(3x^"1 / 2",dx,0,"9 / 4")=6int(x^"1 / 2",dx,0,"9 / 4")}}}
As the antiderivative is {{{int(x^"1 / 2",dx)=(2/3)x^"3 / 2"+C}}} ,
that area is
{{{6int(x^"1 / 2",dx,0,"9 / 4")=6((2/3)(9/4)^"3 / 2"-(2/3)*0^"3 / 2")=6(2/3)(9/4)^"3 / 2"=4((3/2)^2)^"3 / 2"=4(3/2)^3=4*27/8=highlight(27/2=13.5)}}} .
 
ALTERNATIVELY,
you could do like you may have to do in other cases and interchange variables
(or take x as a function of y). 
{{{y^2=9x}}} is equivalent to {{{x=(1/9)y^2}}} taking x as a function of y.
That curve and {{{x=9/4}}} intersect at {{{y=" " +- 9/2}}} .
In other words, the endpoints of the latus rectum are the points with
{{{system(x=9/4,y=" " +- 9/2)}}} .
We can calculate the area between the functions of {{{y}}} ,
{{{x=(1/9)y^2}}} and {{{x=9/4}}} ,
in the interval from {{{y=-9/2}}} to {{{y=9/2}}} as
{{{int((9/4-(1/9)y^2),dy,"-9 / 2","9 / 2")}}} ,
or better yet as
{{{2int((9/4-(1/9)y^2),dy,0,"9 / 2")}}} . As {{{int((9/4-(1/9)y^2),dy)=(9/4)y-(1/9)(y^3/3)}}} ,
the area would be calculated as
{{{2int((9/4-(1/9)y^2),dy,0,"9 / 2")=2"["}}}{{{(9/4)(9/2)-(1/9)(1/3)(9/2)^3-(9/4)*0-(1/9)(0^3/3)}}}{{{"]"=2((9/4)(9/2)-(1/9)(1/3)(9/2)^3)=2(9^2/8-(1/9)(1/3)(9^3/8))=2(9^2/8)(1-1/3)=2*(81/8)(2/3)=highlight(27/2=13.5)}}} .