question_answer

The area of the plane region bounded by the curves \[x+2{{y}^{2}}=0\] and \[x+3{{y}^{2}}=1\] is equal to       AIEEE  Solved  Paper-2007

A) \[\frac{2}{3}\]                   

B)        \[\frac{4}{3}\]   

C)        \[\frac{5}{3}\]                   

D)        \[\frac{1}{3}\]

Correct Answer: B

Solution :

                                \[x+2{{y}^{2}}=0\Rightarrow {{y}^{2}}=-\frac{x}{2}\] parabola \[x+3{{y}^{2}}=1\Rightarrow {{y}^{2}}=-\frac{1}{3}\left( x-1 \right)\] parabola                 Solving equation of two parabolas simultaneously, we get \[x=-2;\,\,y=\pm 1\] Area of the region ABCA \[=\left| \int\limits_{0}^{1}{\left( -2{{y}^{2}}-1+3{{y}^{2}} \right)dy} \right|=\left| \int\limits_{0}^{1}{\left( {{y}^{2}}-1 \right)dy} \right|\]\[=\left| \,\left| \frac{{{y}^{3}}}{3}-y \right. \right|_{0}^{1}\left| =\left| \frac{1}{3}-1 \right|=\frac{2}{3} \right.\] Hence area of region bounded by given curves is equal to \[\frac{4}{3}\].