The area of the region A = \(x, y) : 4x2 + y2 8 and y2 4x\ is:
- + 23
- 2 + 2
- + 4
- 2 + 13
Answer
(1) + 23
Solution
Ellipse: x22 + y28 = 1 (semi-axes 2 and 2 2); parabola: y2 = 4x.
Intersection: 4x2 + 4x = 8
x2 + x - 2 = 0 x = 1 (as x 0), y = 2
[figure]
The region is symmetric about the x-axis, so
A = 2 [ 01 2 x dx + 1 2 2 2 - x2 dx ]
01 2 x dx = 43
1 2 2 - x2 dx = [ x2 2 - x2 + -1 x 2 ]1 2
= 2 - ( 12 + 4 ) = 4 - 12
A = 2 [ 43 + 2 - 1 ]
= + 23
Answer: (1)