The area bounded by the curves `y^(2)=4a(x+a)` and `y^(2)=4b(b-x)`, where `a, bgt0` units

Options

A. `(8)/(3)(a+b)sqrt(ab)` sq unit
B. `(2)/(3)(a+b)sqrt(ab)` sq unit
C. `(2)/(3)(a+b)2sqrt(ab)` sq unit
D. None of the above

Answer

A

Explanation

The two curves are `y^(2)=4a(x+a)" ... (i)"`
and `y^(2)=-4b(x-b)" ... (ii)"`
Clearly, these two curves represents two parabolas having their vertices at (-a, 0) and (b, 0) respectively as shown in figure.

To find teh coordinates of their points of intersection, we solve Eqs. (i) and (ii) together. Solving these equations, we find that the two curves intersect at `A(b-a,2sqrt(ab))` and `B(b-a,-2sqrt(ab))`
`therefore` Required area `=int_(-2sqrt(ab))^(2sqrt(ab))(x_(1)-x_(2))dy`
`=int_(-2sqrt(ab))^(2sqrt(ab)){(b-(y^(2))/(4b))-((y^(2))/(4a)-a)}dy`
`[{:(becauseP(x_(1),y)andQ(x_(2),y)" lie on (ii) and (i) respectively."),(therefore y^(2)=-4b(x_(1)-b)andy^(2)=4a(x_(2)+a)):}]`
`=int_(-2sqrt(ab))^(2sqrt(ab)){a+b-((a+b)/(4ab))y^(2)}dy`
`=2int_(0)^(2sqrt(ab)){a+b-((a+b)/(4ab))y^(2)}dy`
`=2[(a+b)y-((a+b)/(4ab))(y^(3))/(3)]_(0)^(2sqrt(ab))`
`=2[(a+b)2sqrt(ab)-((a+b)/(4ab))(8ab sqrt(ab))/(3)]`
`=2[(a+b)2sqrt(ab)-(2)/(3)(a+b)sqrt(ab)]`
`=(8)/(3)(a+b)sqrt(ab)` sq units