JEE Main & Advanced Sample Paper JEE Main - Mock Test - 24

  • question_answer
    If the chords of contact of tangents from two points \[(-4,2)\] and \[(2,1)\] to the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] are at right angle, then the eccentricity of the hyperbola is

    A) \[\sqrt{\frac{7}{4}}\]    

    B)                    \[\sqrt{\frac{3}{2}}\]           

    C) \[\sqrt{2}\]

    D)        \[\sqrt{\frac{5}{3}}\]

    Correct Answer: B

    Solution :

         [b]   Equation of chord of contact with respect to point \[(-4,2)\] is \[\frac{-4x}{{{a}^{2}}}-\frac{2y}{{{b}^{2}}}=1\]                              ? (1) Equation of chord of contact and with respect to point \[(2,1)\] is \[\frac{2x}{{{a}^{2}}}-\frac{y}{{{b}^{2}}}=1\]                                ? (2) According to question, we have \[\left( \frac{\frac{4}{{{a}^{2}}}}{\frac{-2}{{{b}^{2}}}} \right)\times \left( \frac{\frac{-2}{{{a}^{2}}}}{\frac{-1}{{{b}^{2}}}} \right)=-1\] \[\Rightarrow \,\,\,\frac{{{b}^{4}}}{{{a}^{4}}}=\frac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{1}{2}\] \[\Rightarrow \,\,\,\,\,e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}\]  


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