The area bounded by the curves `y^(2)=4a^(2)(x-1)` and lines x = 1 and y = 4a is

Explanation

The vertex of the parabola ` y^(2) = 4a ^2 ( x - 1 ) ` is A (1, 0) .
To find the points of the intersection of the line ` y = 4a ` and parabola ` y^2 = 4a ^2 ( x - 1 )`.
Put ` y = 4a ` in `y^2 = 4a ^2 ( x - 1 ) `, we get, ` 16 a ^2 = 4a ^2 ( x - 1 ) `
` therefore 4 = x - 1 " " therefore x = 5 `
` therefore ` the point of the intersection is ` C( 5, 4a ) `.
Required area = area of the region ABCDA
= ( area of the rectangle AECD ) - ( area of region AECBA )
Now, area of the rectangle AECD = AE ` xx ` CE = ` 4 xx 4a = 16 a ` sq units.
Area of the region AECBA = area under the parabola ` y^ 2 = 4 a ^2 ( x - 1 ) `
` = int _ 1 ^ 5 y dx `, where ` y = 2a sqrt ( x - 1 ) `
` = int _ 1 ^(5) 2 q sqrt ( x - 1 ) dx = 2a int _ 1 ^( 5a ) ( x - 1 ) ^((1)/(2)) dx `
` = 2a[ ( ( x - 1 ) ^((3)/(2)) ) / ( 3//2) ] _ 1 ^(5) = ( 4a) / ( 3) [ ( x- 1 ) ^((3) /(2)) ] _ 1 ^ ( 5 ) `
` = ( 4a )/ ( 3 ) [ 4^((3)/ ( 2 )) - 0 ]= ( 4a) / ( 3) xx 8 = ( 32a ) / ( 3) ` sq units.
` therefore ` required area ` = 16 a - ( 32 a ) / ( 3) = ( 16 a ) / ( 3 ) ` sq units.