The area of the smaller region bounded by the ellipse and the line is ab/4(π-2).
The area of the smaller region bounded by the ellipse x²/a²+y²/b²=1 and the line x/a+yb=1 can be determined by subtracting the area of the triangle OAB formed by the intersection of the ellipse and the line from the area of the quarter ellipse OABO. The coordinates of point A, where the ellipse and the line intersect, can be found by solving the system of equations:
x²/a²+y²/b²=1
x/a+yb=1
Substituting the second equation into the first equation, we get:
x²/a²+(b/a)²x²=1
Simplifying, we get:
(a²+b²)x²=a²
Solving for x, we get:
x=±a/√(a²+b²)
Since point A is in the first quadrant, we take the positive solution:
x=a/√(a²+b²)
Substituting this value of x back into the second equation, we get:
a/√(a²+b²)+yb=1
Solving for y, we get:
y=b/√(a²+b²)
Therefore, the coordinates of point A are (a/√(a²+b²), b/√(a²+b²)).
The area of triangle OAB is given by:
1/2 * (base * height) = 1/2 * (a/√(a²+b²)) * (b/√(a²+b²)) = ab/2(a²+b²)
The area of the quarter ellipse OABO is given by:
1/4 * πab
Therefore, the area of the smaller region bounded by the ellipse and the line is:
1/4 * πab - ab/2(a²+b²) = ab/4(π-2)