JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    Let a circle of radius \[\frac{2}{\sqrt{3}}\] is touching the lines \[{{x}^{2}}-4xy+{{y}^{2}}=0\]in the first quadrant at points A and B. If area of triangle OAB (0 being the origin) is \[\Delta \] then \[{{\Delta }^{2}}\] equals

    A)  1                    

    B)  3

    C)  2                                

    D)  1

    Correct Answer: B

    Solution :

    Here \[\tan 2\theta \,=\frac{2\sqrt{4}-1}{2}\,=\sqrt{3}\,\Rightarrow \,\theta \,=\frac{\pi }{6}\] \[\therefore \,\,\] Area of \[\Delta OAB=\frac{1}{2}{{(r\cos \theta )}^{2}}(\sin 2\theta )\] \[=\frac{1}{2}\,{{\left( r\sqrt{3} \right)}^{2}}\frac{\sqrt{3}}{2}\,=\frac{3\sqrt{3}{{r}^{2}}}{4}\] \[={{\Delta }^{2}}=3\,\,\left( As\,r=\frac{2}{\sqrt{3}} \right)\]

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