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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Tangent and Normal

  • question_answer
     If the normal to the curve \[{{y}^{2}}=5x-1\], at the point (1, ?2) is of the form \[ax-5y+b=0\], then a and b are                                                               [Pb. CET 2001]

    A)            4, ? 14

    B)            4, 14

    C)            ?4, 14

    D)            ?4, ?14

    Correct Answer: A

    Solution :

               We have, \[{{y}^{2}}=5x-1\]                                            ?..(i)            At \[(1,-2)\]; \[\frac{dy}{dx}={{\left[ \frac{5}{2y} \right]}_{(1,\,-2)}}=\frac{-5}{4}\]            \[\therefore \] Equation of normal at the point (1, ?2) is,                    \[[y-(-2)]\,\left[ \frac{-5}{4} \right]+x-1=0\]            \[\therefore 4x-5y-14=0\]                                                ??(ii)            As the normal is of the form \[ax-5y+b=0\], comparing  this with (ii), we get \[a=4\] and \[b=-14\].


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