If the area of the region bounded by the curves y=x^2 and x=y^2 is k, then the area of the region bounded by the curves x+ 3y2= ( 3x-y2 )^2 and 3x-y2= ( x+ 3y2 )^2, is

Question

Accepted Answer

Correct answer: C. (c) Given, Curves are x+ 3y2= ( 3x-y2 )^2 and 3x-y2= ( x+ 3y2 )^2 Let x+ 3y2= x+ 3y 1+ ( 3 )^2=y and 3x-y2= 3x-y ( 3 )^2+1=x so, the given curve x+ 3y2= ( 3x-y2 )^2, becomes y=x^2.......(i) Similarly, the given curve 3x-y2= ( x+ 3y2 )^2 becomes x^2=y......(ii) So the area of the region bounded by curves (i) and (ii) is equals to the area of the region bounded by the curve y=x^2 and x=y^2 is equal to ' k ' (given).

$If the area of the region bounded by the curves y = x^{2} and x = y^{2} is k, then the area of the region bounded by the curves \frac{x + 3 y}{2} = (\frac{3 x - y}{2})^{2} and \frac{3 x - y}{2} = (\frac{x + 3 y}{2})^{2}, is$

Solution

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