Question

Find the area enclosed by the parabola 4 y = 3 x 2 and the line 2 y = 3 x + 12

Answer 1

We have to find the area enclosed by the parabola whose equation is 4y=3 x 2 , and the line 2y=3x+12 . Draw the graphs of the equations and shade the common region.

Figure (1)

Solve the equation of parabola and straight line to find the points of intersection.

2y=3x+12 y= 3x 2 +6

Substitute this value of y in the equation of parabola.

4( 3x 2 +6 )=3 x 2 6x+24=3 x 2 3 x 2 −6x−24=0 x 2 −2x−8=0

Further, solve the above equation.

x 2 −4x+2x−8=0 ( x−4 )x+2( x−4 )=0 ( x−4 )( x+2 )=0 x=−2,4

Corresponding values of y are,

y= 3 4 x 2 , 3 4 x 2 = 3 4 ( −2 ) 2 , 3 4 ( 4 ) 2 =3,12

Since the point of intersection is known, drop two perpendiculars from the points of intersection to the x-axis.

The area of the region OBAO is,

Area of the region OBAO=Area of the region CABD−Area of the region CAOBDC

To find the area bound by the straight line 2y=3x+12 with the x-axis, assume a vertical strip of infinitesimally small width and integrate the area.

Area of the region CABD= ∫ −2 4 ( 3x 2 +6 ) dx = [ 3 x 2 4 +6x ] −2 4 =[ 3 ( 4 ) 2 4 +6( 4 )−( 3 ( −2 ) 2 4 +6( −2 ) ) ] =45

Similarly, find the area bound by the parabola with x-axis,

Area of the region CAOBDC= ∫ −2 4 y dx

From the equation of parabola, find the value of y in terms of x and substitute in the above integral.

Area of the region CAOBDC= ∫ −2 4 3 4 x 2 dx = 3 4 [ x 3 3 ] −2 4 = 1 4 [ ( 4 ) 3 − ( −2 ) 3 ] = 1 4 ( 64+8 )

Further, solve the above equation.

Area of the region CAOBDC= 1 4 ( 72 ) =18

Area of the region OBAO=Area of the region CABD−Area of the region CAOBDC =45−18 =27

Thus, the area enclosed by the parabola whose equation is 4y=3 x 2 , and the line 2y=3x+12 is 27sq units .