Home
Class 12
MATHS
If the area (in sq. units) bounded by th...

If the area (in sq. units) bounded by the parabola `y^(2) = 4 lambda x` and the line `y = lambda x, lambda gt 0`, is `(1)/(9)`, then `lambda` is equal to:

A

`2sqrt(6)`

B

48

C

24

D

`4sqrt(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given , equation of curves are
`y^(2)=4lambda x " …(i)" `
`and y = lalmbda x " …(ii)" `
`lambda gt 0`
Area bounded by above two curve is, as per figure

the intersection point A we will get on the solving Eqs. (i) and (ii) we get
`lambda^(2)x^(2)=4lambda x`
`rArr x=(4)/(lambda), " so " y=4`
So, `A((4)/(lambda),4)`
Now, required area is
`=int_(0)^(4//lambda)(2sqrt(lambda x)-lambdax)dx`
`=2sqrt(lambda)[(x^(3//2))/((3)/(2))]_(0)^(4//lambda)-lambda[(x^(2))/(2)]_(0)^(4//lambda)`
`=(4)/(3)sqrt(lambda)(4sqrt(4))/(lambda sqrt(lambda))-(lambda)/(2)((4)/(lambda))^(2)`
`=(32)/(3lambda)-(8)/(lambda)=(32-24)/(3lambda)=(8)/(3lambda) `
It is given that area `=(1)/(9)`
`rArr (8)/(3lambda)=(1)/(9)`
`rArr lambda =24`
Promotional Banner

Similar Questions

Explore conceptually related problems

If the area (in sq.units) bounded by the parabola y^(2)=4 lambda x and the line y=lambda x,lambda>0, is (1)/(9), then lambda is equal to:

If x+y+1 = 0 touches the parabola y^(2) = lambda x , then lambda is equal to

if the area enclosed by the curves y^(2)=4 lambda x and y=lambda x is (1)/(9) square units then value of lambda is

The area bounded by y=sqrt(1-x^(2)) and y=x^(3)-x is divided by y - axis in the ratio lambda:1(AA lambda gt 1) , then lambda is equal to

The area of the region (in sq units), in the first quadrant, bounded by the parabola y = 9x^(2) and the lines x = 0, y = 1 and y = 4, is

If the line y=2x+lambda be a tangent to the hyperbola 36x^(2)-25y^(2)=3600 , then lambda is equal to

The angle between x^(2)+y^(2)-2x-2y+1=0 and line y=lambda x + 1-lambda , is

If (x+y)^(2)=2(x^(2)+y^(2)) and (xy+lambda)^(2)=4,lambda>0, then lambda is equal to