----------------*---------/--------x
The parabolas are oriented in opposition, but one is displaced some distance along the X axe.
As the figure has some simmetry we can focus just on one of the parabolas, We consider it center in origen. So its equation si
y=ax^2 (a to be determined)
O (x1,y1) ,Q (x2,y2) are the points of interseccion.
Angle z is the angle between tangent on Q and X axe.
We have four main relations:
From y' 2ax2 = tgz => x2=tgz /2a | x1=-1/(2a*tgz) (1)
x1+x2 = 3 sen z + 4 cos z (2)
y2-y1 = 4 sen z - 3 cos z (3)
(y2-y1) = a (x2^2-x1^2) (4)
Everything here is easily express as f(z)
From (1) and (2) is easy to get "a" as f(z)
a = 1/[2 sen z cos z (4 cos z + 3 sen z)]
From (3) and (4) and replacing a=f(z) we get the equation:
3 sen^3 z - 4 cos^3 z = 2 sen z cos z (4 sen z - 3 cos z)
This cubic has solution for z=53,13° (sen z=4/5; cos z=3/5; tg z=4/3 :
(this last value shows that the diagonal of the 3x4 rectangle is parallel to Y axe. If we had supposed this we would have had immediately tg z = 4/3)
(other real solution is for 106,26° and immaginary for i45°)
Then a=125/576
(x1, y1) = (-216/125, 81/125) and (x2, y2) = (384/125, 768/375)
The requested area is enclosed into a rectangle of dimensions (x1+x2 and y1+y2). The area of this rectangle is 12,94
Now we can integrate between x2 and x1 to obtain the area under the parabola not inclused in the interseccion of the parabolas.
(ax^3)/3 ] between (x1,x2) = 2.47
In the zone above we have to exclused another equal area.
So the requested area is 12,94 - 2,47*2 = 8
Harder way?
