11. The area bounded by the parabola y = 4ax and x2 = 4ay is
16a
32a
(a)
(b)
(c)
3
3
(d)
3
3
8a
64a?
3
Answers
Aɴsᴡᴇʀ࿐
Let Z be the number
Z = 13x + 11 where x is the quotient when Z is divided by 13
Z = 17y + 9 where y is the quotient when Z is divided by 17
13x + 11 = 17y + 9
13x + 2 = 17y since x and y are quotients they should be whole numbers . Since y has to be a whole number the left hand side should be multiple of 17
The least possible value of x satisfying the condition is 9 and y will be 7
The answer is 13*9 + 11 = 128 or it is 17*7 + 9 = 128
This is the least number possible. There will be multiple answers and will increase in multiples 17*13 = 221 like 349 , 570, etc.
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Answer:
= Area of the shaded region
Hence, Area = 16a ^2/ 3
Step-by-step explanation:
To Prove that the area enclosed between two parabolas y
2 =4ax and x 2 =4ay
is 3^ 16a/ 3
Given curves are
y ^2 =4ax and x ^2 =4ay
First we have to find the area of Intersection of the two curves
Point of Intersection of the two curves are
( 4a x 2 )
2 =4ax
( 16a 2/ x 4 )=4ax
x 4 =64a 3 x
x 4 −64a 3 x=0
x(x 3 −64a 3 )=0
x=0,x=4a
Also y=0,y=4a
The Point of Intersection of these 2 curves are (0,0) and (4a, 4a )
The Area of the two region between the two curves
= Area of the shaded region
Hence , Area = 16a ^2/ 3