BITSAT Mathematics Question (2024) — Solution

Step-by-step solution

We have given, y=a x^2 ....(i) and x=a y^2 ......(ii) Put the value of y by Eq. (i) in Eq. (ii), we get x=a a^2 x^4 x^4 a^3-x=0 x (x^3 a^3-1 )=0 x=0, 1 2 When, x=0 y=0 and x= 1 a y= 1 a Here, points of intersection of curves y=a x^2 and x=a y^2 are (0,0) and ( 1 a , 1 a ). Required area A= _ x=a ^ x=b [f_2(x)-f_1(x) ] d x 3= _0^ 1 / a ( x a -a x^2 ) d x 3= [ 1 a 2 3 x^ 3 / 2 - a x^3 3 ]_0^ 1 / a 3= 2 3 a [ ( 1 a )^ 3 / 2 ]- a 3 [ ( 1 a )^3 ] 3= 2 3 a 1 a a - a 3 1 a^3 3= 2 3 a^2 - 1 3 a^2 3= 2-1 3 a^2 9 a^2=1 a^2= 1 9 a= 1 3