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

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Question 1086398: find the area bounded by the curve x^2=8y and its latus rectum
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
The equation of the parabola can be written +y+=+%281%2F8%29x%5E2+
Graphed below (green line is latus rectum):
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+graph%28+400%2C+400%2C+-4%2C+4%2C+-5%2C+5%2C+%281%2F8%29%2Ax%5E2%2C+y=2%29%0D%0A+
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For a (vertical) parabola with vertex @(h,k): +4p%28y-k%29+=+%28x-h%29%5E2+
Vertex is at (0,0) so h=k=0: so this parabola has equation +y+=+%281%2F8%29x%5E2+
Solve for p (the distance from vertex to focus) by comparison: +y+=+x%5E2%2F%284p%29+ ==> +4p+=+8+ ==> p=2
So focus is at (0,2). Latus rectum passed through (0,2) and intersects parabola where y+=+2+=+%281%2F8%29x%5E2+ or (-4,2) and (4,2)
Using just the first quadrant & symmetry with 2nd quadrant:
Area = +2%2A+int%28%28+2+-+%281%2F8%29x%5E2%29%2C+dx%2C+0%2C+4%29+
= +2%2A+%282x+-+%281%2F24%29x%5E3%29+ evaluated at 4 and 0
= +2%2A%28%288-%281%2F24%29%2864%29%29+-+%280-0%29%29+
= +highlight%2832%2F3%29+