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The area enclosed between the curves y=k...

The area enclosed between the curves `y=kx^2" and " x=ky^2`,`(kgt0)` is 1 sq. units. Then k is

A

`sqrt(3)`

B

`(1)/(sqrt(3))`

C

`(2)/(sqrt(3))`

D

`(sqrt(3))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
We know that, area of region bounded by the parabolas
` x^(2)=4ay and y^(2)=4bx " is " (16)/(3)(ab)` sq units.
On comparing `y=kx^(2) and x=ky^(2)` with above equations,
we get `4a=(1)/(k) and 4b=(1)/(k)`
`rArr a=(1)/(4k) and b=(1)/(4k)`
` therefore " Area enclosed between " y=kx^(2) and x=ky^(2)` is
` (16)/(3)((1)/(4k))((1)/(4k))=(1)/(3k^(2))`
`rArr (1)/(3k^(2))=1 ` [given, area=1 sq.units]
`rArr k^(2)=(1)/(3) rArr k pm (1)/(sqrt(3))`
` rArr k=(1)/(sqrt(3)) " "[because k gt 0]`
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