The area enclosed between the parabola y 2 = 4 a x y^2=4ax y 2 = 4 a x , and the line y = m x y=mx y = m x , is represented by the shaded areaO A B O OABO O A BO as
The points of intersection of both the curves are( 0 , 0 ) (0,0) ( 0 , 0 ) and( 4 a m 2 , 4 a m ) (\dfrac{4a}{m^2},\dfrac{4a}{m} ) ( m 2 4 a , m 4 a ) .
We draw AC perpendicular to x − a x i s x-axis x − a x i s .
So, A r e a O A B O = A r e a O C A B O − A r e a ( O C A ) Area OABO= Area OCABO - Area (OCA) A re a O A BO = A re a OC A BO − A re a ( OC A )
= ∫ 0 4 a m 2 2 a x − = \int_0^{\dfrac{4a}{m^2}}2\sqrt{ax}- = ∫ 0 m 2 4 a 2 a x − ∫ 0 4 a m 2 m x d x \int_0^{\dfrac{4a}{m^2}}mxdx ∫ 0 m 2 4 a m x d x
=2 a [ x 1.5 1.5 ] 0 4 a m 2 2\sqrt{a}[\dfrac{x^{1.5}}{1.5}]_0^\dfrac{4a}{m^2} 2 a [ 1.5 x 1.5 ] 0 m 2 4 a − - − m [ x 2 2 ] 0 4 a m 2 m[\dfrac{x^2}{2}]^{\dfrac{4a}{m^2}}_0 m [ 2 x 2 ] 0 m 2 4 a
= 8 a 2 3 m 3 u n i t s =\dfrac{8a^2}{3m^3} units = 3 m 3 8 a 2 u ni t s
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