• # question_answer The parabolas ${{y}^{2}}=4x$ and ${{x}^{2}}=4y$ divide the square region bounded by the lines $x=4$, $y=4$and the coordinate axes. If ${{S}_{1}},{{S}_{2}},{{S}_{3}}$ are respectively the areas of these parts numbered from top to bottom, then ${{S}_{1}}:{{S}_{2}}:{{S}_{3}}$ is                                                            [AIEEE 2005] A)            $2:1:2$                                  B)            $1:1:1$                                  C)            $1:2:1$                                  D)            $1:2:3$

${{y}^{2}}=4x$ and ${{x}^{2}}=4y$ are symmetric about line $y=x$                    Þ Area bounded between ${{y}^{2}}=4x$ and $y=x$ is $\int_{0}^{4}{(2\sqrt{x}-x)dx=\frac{8}{3}}$                                       Þ  ${{A}_{s}}_{_{2}}=\frac{16}{3}$ and ${{A}_{{{s}_{1}}}}={{A}_{{{S}_{3}}}}=\frac{16}{3}$                    Þ ${{A}_{{{S}_{1}}}}:{{A}_{{{S}_{2}}}}:{{A}_{{{S}_{2}}}}::1:1:1$.