The area cut off a parabola `4y=3x^(2)` by the straight line `2y=3x+12` in square units, is

Options

A. `27` sq units
B. `12` sq units
C. `33` sq units
D. `21` sq units

Answer

A

Explanation

Clearly `4y=3x^2` is an upward parabola with its vertex at (0,0) And 3x- 2y+12=0 is aline The given equations are
`4y = 3x^2`
and 3x-2y+12=0
The points of intesection of the given parabola and the given line will be obtined by solving (i) and
(ii) simultaneously
Putting `y=3/4 x^2` from (i) in, (ii) ltbegt We get
`3x-3/2 x^2 +12 =0 rArr x^2 + 12 =0 `
The points of intersection of the given parabola and the given line will be obtioned by solving (i) and (simulataneously
Putting `y= 3/4 x^2 ` from (i) in (ii)
We get
`3x - 3/2 x^3 + 12 =0 rArr x^2- 2x-8=0`
`rArr x^2 -4x + 2x -8 =0`
`rArr x(x-4) + 2 (x-4)=0`
`rArr (x-4)(x+2)=0 `
`rArr x= (-2 or x=4`
Now , `(x=-2 rArr y =3) "and " (x = 4 rArr y= 12).`
So , the points of intersection of ( i) and (ii) are A(-2,3) and B( 4,12)
Draw `A L_|_OX "and " BM _|_ OX.`
Required aera= area ALMBA )-(area ALOMBOA)
`=underset(-2)overset(4)int"(y of the line )"dx - underset(-2)overset(4)int"(y of the parabola)"dx`
`=underset(-2)overset(4)int((3x+12))/2dx - underset(-2)overset(4)int(3x^2)/(4)dx=[(3x^2)/(4)+6x]_-2^4-[(x^3)/4]_-2^4`
`=(45 -18 )= 27` sq units
Hence the required area is 27 sq units