SOLUTION: find the area bounded by the curve y^2=9x and its latus rectum

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Question 1110815: find the area bounded by the curve y^2=9x and its latus rectum
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
y%5E2=9x
In this case, the latus rectum is part of the line x=9%2F4=2.25 .
Parabola, focus, directrix and latus rectum look like this
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%3E=0 ,
y%5E2=9x is equivalent to y=3sqrt%28x%29=3x%5E%221+%2F+2%22 , so area=2
As the antiderivative is int%28x%5E%221+%2F+2%22%2Cdx%29=%282%2F3%29x%5E%223+%2F+2%22%2BC ,
that area is
.

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%5E2=9x is equivalent to x=%281%2F9%29y%5E2 taking x as a function of y.
That curve and x=9%2F4 intersect at y=%22+%22+%2B-+9%2F2 .
In other words, the endpoints of the latus rectum are the points with
system%28x=9%2F4%2Cy=%22+%22+%2B-+9%2F2%29 .
We can calculate the area between the functions of y ,
x=%281%2F9%29y%5E2 and x=9%2F4 ,
in the interval from y=-9%2F2 to y=9%2F2 as
int%28%289%2F4-%281%2F9%29y%5E2%29%2Cdy%2C%22-9+%2F+2%22%2C%229+%2F+2%22%29 ,
or better yet as
2int%28%289%2F4-%281%2F9%29y%5E2%29%2Cdy%2C0%2C%229+%2F+2%22%29 . As int%28%289%2F4-%281%2F9%29y%5E2%29%2Cdy%29=%289%2F4%29y-%281%2F9%29%28y%5E3%2F3%29 ,
the area would be calculated as
2int%28%289%2F4-%281%2F9%29y%5E2%29%2Cdy%2C0%2C%229+%2F+2%22%29=2%22%5B%22 .