SOLUTION: find the area bounded by the curve x^2=8y and its latus rectum

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the area bounded by the curve x^2=8y and its latus rectum       Log On


   



Question 1086384: find the area bounded by the curve x^2=8y and its latus rectum
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
The equation for a vertical parabola in terms of the vertex and focus is:
4p(y-k) = (x-h)^2, where (h,k) is the vertex and p is the distance from vertex to focus
In this case, the vertex is (0,0), and we can read off the value of p:
x^2 = 4py -> p = 2
So we need to find the area between the line y = 2 and the curve y = x^2/8
The area will be the difference of the functions integrated from -4 to 4,
since these are the endpoints where the two curves meet.
A =
A = 2*(2*4 - 64/24) = 32/3