The area (in sq units) enclosed by the loop of the curve (a y^2=x^2(a-x),(a>0) ) is

Step-by-step solution

Given curve is (a y^2=x^2(a-x),(a > 0) ) The required area ( aligned & =2 ₀^a x a-x a d x & = 2 a ₀^a x a-x d x aligned ) Put (a-x=t^2 ), at (x=0, t= a ) and , at (x=a, t=0 ) and (-d x=2 t d t ) So, required area (=- _ a ^0 4 a t^2 (a-t^2 ) d t ) ( aligned & = 4 a ₀^ a (a t^2-t^4 ) d t= 4 a [ a t^3 3 - t^5 5 ]₀^ a & = 4 a ( a^2 a 3 - a^2 a 5 )= 8 15 a^2 aligned ) Hence, option (d) is correct.