Find the area enclosed between the parabola `y^2=4a x` and the line `y = m x`.

Explanation

The given equation are
`y^2=4ax " "…(i)`
and y= mx ….(ii) Clearly `y^2=4ax ` is a right handed parabola passing through the origin And y=mx is a line passing through the origin .
In order to find points of intersection of the given parabola and putting y=mx from (ii) and (ii) simultaneously
Putting y=mx from (ii) into (i) we get
`m^2 x^2 = 4ax rArr x(m^2 x 4a)=0`
`m^2 x^2j = 4ax rArr x(m^2x-4a)=0`
`rArr x =0 or x=4a/m^2`
Now `(x =0 rArr y= 0 ) and (x =(4a)/(m^2)rArr y = (4a)/(m))`
So , the points of intersection of the given parabola and the chord are
`O(0,0) and A ((4a)/(m^2),rArr y= (4a)/(m))`
Draw ` AM _|_ OX`
Required area = (area OBAMO)=(area OAMO)
`=underset(0)overset(4a//m^2)int`(y for the parabola ) dx-` =underset(0)overset(4a//m^2)int `(y for the line )dx
`=underset(0)overset(4a//m^2)int 2 sqrt(ax)dx - =underset(0)overset(4a//m^2)int mx dx `
`=2sqrt(a).""2/3[x^(3//2)]_0^(4a//m^2)-[(mx^2)/(2)]_0^(4a//m^2)`
`=[(4sqrt(2))/(3).""8/m^3a^(3//2)-m/2.""(16a^2)/(m^4)]`
`=((32a^2)/(3m^2)-(8a^2)/(m^3))=((8a^2)/(3m^2))`sq units
Hence , the required area is `((8a^2)/(3m^2))` sq units .