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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    If \[5x+9=0\]is the directrix of the hyperbola \[16{{x}^{2}}9{{y}^{2}}=144,\]then its corresponding focus is: [JEE Main 10-4-2019 Afternoon]

    A) \[\left( -\frac{5}{3},0 \right)\]             

    B) \[(5,0)\]

    C) \[(-5,0)\]

    D) \[\left( \frac{5}{3},0 \right)\]

    Correct Answer: C

    Solution :

                \[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{16}=1\] \[a=3,b=4\And e=\sqrt{1+\frac{16}{9}}=\frac{5}{3}\] corresponding focus will be (-ae, 0) i.e., (-5, 0).          


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