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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    Let a and b respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation \[\text{9e}{{\text{-}}^{\text{2}}}\text{-18e}+\text{5}=0.\] If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of hyperbola, then \[{{a}^{2}}-{{b}^{2}}\]is equal to   JEE Main Online Paper (Held On 09 April 2016)

    A) - 7          

    B) - 5

    C) 5                                             

    D) 7

    Correct Answer: A

    Solution :

                    \[9{{e}^{2}}-18e+5=0\]\[\Rightarrow \]\[e=\frac{5}{3}\]                 \[\therefore \]\[1+\frac{{{b}^{2}}}{{{a}^{2}}}={{e}^{2}}=\frac{25}{9}\]                                    ?(i) Also distance between foci and directrix is \[=\left( ae-\frac{a}{e} \right)=5-\frac{9}{5}\] \[\Rightarrow \]\[a\left( \frac{5}{3}-\frac{3}{5} \right)=\frac{16}{5}\Rightarrow a=3\]from (i) \[1+\frac{{{b}^{2}}}{9}={{e}^{2}}=\frac{25}{9}\Rightarrow {{b}^{2}}=16\] \[\therefore \]\[{{a}^{2}}-{{b}^{2}}=9-16=-7\]                                


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