JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    What will be equation of that chord of hyperbola \[25{{x}^{2}}-16{{y}^{2}}=400\], whose mid point is (5, 3) [UPSEAT 1999]

    A)            \[115x-117y=17\]                     

    B)            \[125x-48y=481\]

    C)            \[127x+33y=341\]                     

    D)            \[15x+121y=105\]

    Correct Answer: B

    Solution :

                      According to question, \[S\equiv \,25{{x}^{2}}-16{{y}^{2}}-400=0\]            Equation of required chord is \[{{S}_{1}}=T\]           .....(i)            Here, \[{{S}_{1}}=25{{(5)}^{2}}-16{{(3)}^{2}}-400\]                    \[=625-144-400=81\]            and \[T\equiv 25x{{x}_{1}}-16y{{y}_{1}}-400,\] where \[{{x}_{1}}=5,\,{{y}_{1}}=3\]            \[=25(x)(5)-16(y)(3)-400\]\[=125x-48y-400\]            So from (i), required chord is                    \[125x-48y-400=81\] or \[125x-48y=481.\]


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