A) 2
B) \[-2\]
C) \[\frac{1}{2}\]
D) None of these
Correct Answer: A
Solution :
The two curves \[{{y}^{2}}=4ax\] and \[y=mx\] intersect at \[\left( \frac{4a}{{{m}^{2}}},\frac{4a}{m} \right)\] and the area enclosed by the two curves is given by \[\int_{0}^{4a/{{m}^{2}}}{(\sqrt{4ax}-mx)}\,dx\]. \[\therefore \,\,\,\int_{0}^{4a/{{m}^{2}}}{(\sqrt{4ax}-mx)}\,dx=\frac{{{a}^{2}}}{3}\] Þ \[\frac{8}{3}\frac{{{a}^{2}}}{{{m}^{3}}}=\frac{{{a}^{2}}}{3}\Rightarrow {{m}^{3}}=8\Rightarrow m=2\].
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