• # question_answer The area bounded by the parabolas ${{y}^{2}}=4a(x+a)$and ${{y}^{2}}=-4a(x-a)$is A)  $\frac{16}{3}{{a}^{2}}\text{sq units}$ B)         $\frac{8}{3}\text{sq}\,\text{units}$     C)         $\frac{\text{4}}{\text{3}}{{\text{a}}^{\text{2}}}\text{sq}\,\text{units}$ D)         None of these

The focus of the parabolas ${{y}^{2}}=4a(x+a)$and${{y}^{2}}=-4a(x-a)$  are  $(-a,0)$   and   $(a,0)$ respectively. The required area is shown in the figure which is given by as following. Area of $ABCD=2.$area of $AOCB$                                                 $=4.$area of AOB                                                 $=4.\int_{-a}^{0}{\sqrt{4a(x+a)}}dx$                                                 $=4.2\sqrt{a}\int_{-a}^{0}{\sqrt{x+a}}\,dx$                                                 $=8\sqrt{a}\left[ \frac{2}{3}{{(x+a)}^{3/2}} \right]_{-a}^{0}$                 $\Rightarrow$Area of $ABCD=8\sqrt{a}.\frac{2}{3}({{a}^{3/2}}-0)$                                                 $=\frac{16}{3}\sqrt{a}.{{a}^{3/2}}$                                                        $=\frac{16}{3}{{a}^{2}}$                 which is the required area. You will be redirected in 3 sec 