Question

Find the area enclosed between the parabola y 2 = 4 ax and the line y = mx

Answer 1

We have to find the area enclosed by the parabola whose equation is y 2 =4ax , and the line y=mx . 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.

y 2 =4a( y m ) m y 2 −4ay=0 y( my−4a )=0 y=0, 4a m

Corresponding values of x are,

x= y m =0, 4a m 2

Since the point of intersection is known, drop a perpendicular from the point of intersection to the x-axis.

The area of the region OBAO is,

Area of the region OBAO=Area of the region OBACO−Area of the triangle OAC

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

Area of the triangle OAC= ∫ 0 4a m 2 mx dx = [ m x 2 2 ] 0 4a m 2 =[ m ( 4a m 2 ) 2 2 −0 ] = 8 a 2 m 3

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

Area of the region OBACO= ∫ 0 4a m 2 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 OBACO= ∫ 0 4a m 2 2 ax dx =2 a ∫ 0 4a m 2 x dx =2 a [ ( x ) 1 2 +1 1 2 +1 ] 0 4a m 2 =2⋅ 2 3 a [ x 3 2 ] 0 4a m 2

Further, solve the above equation.

Area of the region OBACO= 4 3 a ( 4a m 2 ) 3 2 = 32 a 2 3 m 3

Area of the region OBAO=Area of the region OBACO−Area of the triangle OAC = 32 a 2 3 m 3 − 8 a 2 m 3 = 8 a 2 3 m 3

Thus, the area enclosed by the parabola whose equation is y 2 =4ax , and the line y=mx is 8 a 2 3 m 3  sq units .