Two parabolas intersect at two points. The four tangent lines at their points of intersection form a 3 * 4 rectangle. Find the area of the closed region created by both parabolas.
The problem implies the area is fixed by the size of the rectangles. Assuming this is so we can rotate and translate the rectangle so one corner is on the origin and one side has slope 4/3. The opposite corner is some point (p,q)
One of the parabolas has an equation of the form
f(x)=ax^2+bx+c, f'(x)=2ax+b
and it is know
f(0)=0
f'(0)=4/3
from which we can easily derive c=0, b=4/3
but also
f(p)=q
f'(p)=-4/3
-4/3=2ap+4/3
p=-25/(24a)
q=ap^2+4/3p
q=-175/576a
Since (p,q) is 5 units from the origin we have
p^2+q^2=5^2
solve to get a=-125/576
and (p,q)=(24/5,7/5)
From here it's just a little integral calculus to get the area of the parabola inside half of the square = 4
and double this to get the area inside the two parabolas = 8
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Posted by Jer
on 2018-03-25 11:59:35 |
Hard way
