question_answer

The area enclosed by the parabola \[{{y}^{2}}=4ax\] and the straight line \[y=2ax,\] is                                    [MP PET 1993]

A)            \[\frac{{{a}^{2}}}{3}\] sq. unit                                          

B)            \[\frac{1}{3{{a}^{2}}}\] sq. unit

C)            \[\frac{1}{3a}\] sq. unit   

D)            \[\frac{2}{3a}\] sq. unit

Correct Answer: C

Solution :

           The points of intersection of the parabola \[{{y}^{2}}=4ax\]and the chord\[y=2ax\]is obtained by solving these equations simultaneously.            \[{{y}^{2}}=4ax,y=2ax\Rightarrow {{(2ax)}^{2}}=4ax\]            Þ \[x[4{{a}^{2}}x-4a]=0\]\[\Rightarrow 4ax[ax-1]=0\]            Þ \[x=0\]or \[x=\frac{1}{a}\]Also \[x=0\Rightarrow y=0\]and \[x=\frac{1}{a}\]            Þ\[y=\pm 2\]                    Hence the required points are (0,0) and \[\left[ \frac{1}{a},2 \right]\].                    Now required area\[=\int_{0}^{1/a}{[\sqrt{4ax}-2ax}]dx=\frac{1}{3a}sq\]. unit.