A) \[\pm \,3\sqrt{2}\]
B) \[\pm \,\frac{2}{\sqrt{3}}\]
C) \[\sqrt{2}\]
D) \[\sqrt{3}\]
Correct Answer: B
Solution :
Given curves; \[y=m{{x}^{2}}\] and \[{{y}^{2}}m=x;\,m>0\] Intersection point of both curves; \[x=m{{(m{{x}^{2}})}^{2}}={{m}^{3}}{{x}^{4}}\] \[\Rightarrow \] \[{{m}^{3}}{{x}^{4}}-x=0\] \[\Rightarrow \] \[x({{m}^{3}}{{x}^{3}}-1)=0\] \[\Rightarrow \] \[x(mx-1)({{m}^{2}}{{x}^{2}}+1+mx)=0\] \[\Rightarrow \] \[x=0,x=1/m\] and \[y=0,y=1/m\] We take only the points \[=(0,0)\] and \[(1/m,1/m)\] Now, the area of the curve \[=\int_{0}^{1/m}{\left( \sqrt{\frac{x}{m}}-m{{x}^{2}} \right)}\,dx\] Given, \[\frac{1}{4}=\left[ \frac{2}{3\sqrt{m}}.{{x}^{3/2}}-m.\frac{{{x}^{3}}}{3} \right]_{0}^{1/m}\] \[\Rightarrow \] \[\frac{1}{4}=\left[ \frac{2}{3\sqrt{m}}.\frac{1}{{{m}^{3/2}}}-\frac{m}{3}.\frac{1}{{{m}_{3}}} \right]\] \[\Rightarrow \] \[\frac{1}{4}=\left\{ \frac{2}{3{{m}^{2}}}-\frac{1}{3{{m}^{2}}} \right\}\] \[\Rightarrow \] \[\frac{1}{4}=\frac{1}{3{{m}^{2}}}\Rightarrow {{m}^{2}}=\frac{4}{3}\] \[\therefore \] \[m=\pm \,\frac{2}{\sqrt{3}}\]
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