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The area enclosed between the curves `y=ax^2 and x=ay^2(a > 0)` is 1 sq. unit, value of a is `1/sqrt3` (b) `1/2` (c) 1 (d) `1/3` |
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Answer» Equation of the given two curves are, `y = ax^2, x = ay^2` Intersection points of these two curves will be, `(1/a,1/a)` and `(0,0)`. So, area between these two curves will be, `int_0^(1/a)(sqrt(x/a) - ax^2) dx = 1` `=>[1/sqrta x^(3/2)/(3/2) - (ax^3)/3]_0^(1/a) = 1` `=>2/3*1/a^2 - 1/3*1/a^2 = 1` `=>1/a^2 = 3` `=>a^2 = 1/3` `=> a= 1/sqrt3` So,option-`(a)` is the correct option. |
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