If the area of the region bounded by 16x2-9y2=144 and 8x-3y=24 is A, then 3 (A+6 e(3) ) is equal to .
Answer
24
Solution
Hyperbola: x29- y216=1; line: y= 8x-243
Intersection: 16x2-(8x-24)2=144 x2-8x+15=0 x=3, 5
Points: (3, 0) and (5, 163 )
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For 3 x 5 the upper branch y= 43 x2-9 lies above the line.
A= 35 ( 43 x2-9- 8x-243 )dx
35 x2-9 dx= [ x2 x2-9- 92 e |x+ x2-9 | ]35
= (10- 92 e 9 )- (0- 92 e 3 )=10- 92 e 3
35 8x-243 dx= 13 [4x2-24x ]35= 13(-20+36)= 163
A= 43 (10- 92 e 3 )- 163=8-6 e 3
3 (A+6 e 3 )=3 8=24
Answer: 24