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JEE Main & Advanced Sample Paper JEE Main Sample Paper-10

  • question_answer
    The tangent at a point on the hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]meets one of its directrix in F. If PF subtends an angle \[\theta \] at the corresponding focus, then \[\theta \] equals

    A)  \[3\pi /4\]                         

    B)  \[\pi /2\]

    C)  \[3\pi /2\]                         

    D)  \[\pi /6\]

    Correct Answer: B

    Solution :

     Let directrix be x = ale and the focus be S (ae, 0). Let P(a sec \[\text{ }\!\!\theta\!\!\text{ }\], b tan \[\text{ }\!\!\theta\!\!\text{ }\]) be any point on the curve. Equation of tangent at P is \[\frac{\text{x}\,\text{sec}\,\text{ }\!\!\theta\!\!\text{ }}{a}-\frac{y\,\tan \,\text{ }\!\!\theta\!\!\text{ }}{b}=1\] Let F be the intersection point of the tangent and the directrix, so that \[F\equiv \left( \frac{a}{e},\frac{b(\sec \theta -e)}{e\tan \theta } \right)\] \[\Rightarrow \]               \[{{m}_{SF}}=\frac{b(\sec \theta -e)}{-a\,\tan \theta ({{e}^{2}}-1)},\]                 \[{{m}_{PS}}=\frac{b\tan \theta }{a(\sec \theta -e)}\] \[\Rightarrow \]               \[{{m}_{SF}}.{{m}_{ps}}=-1\]                      \[\theta =\frac{\pi }{2}\]


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