question_answer

If the area bounded by the parabolas \[{{y}^{2}}=4\alpha (x+\alpha )\] and \[{{y}^{2}}=-4\alpha (x-\alpha )\], where \[\alpha >0\] is 48 square units then \[\alpha \] is equal to

A)  \[\frac{1}{3}\]                                                  

B)  \[\frac{1}{9}\]

C)  3                                            

D)  9

Correct Answer: C

Solution :

Area \[=\frac{16}{3}{{\alpha }^{2}}=48\Rightarrow \alpha =3\]