JEE Main & Advanced Mathematics Conic Sections Question Bank Self Evaluation Test - Conic Sections

  • question_answer
    If \[x=9\] is the chord of contact of the hyperbola \[{{x}^{2}}-{{y}^{2}}=9\], then the equation of the corresponding pair of tangents is

    A) \[9{{x}^{2}}-8{{y}^{2}}+18x-9=0\]

    B) \[9{{x}^{2}}-8{{y}^{2}}-18x+9=0\]

    C) \[9{{x}^{2}}-8{{y}^{2}}-18x-9=0\]

    D) \[9{{x}^{2}}-8{{y}^{2}}+18x+9=0\]

    Correct Answer: B

    Solution :

    [b] The equation of tangent at a point \[({{x}_{1}},{{y}_{1}})\] on the hyperbola \[{{x}^{2}}-{{y}^{2}}=c\] is given by \[x{{x}_{1}}-y{{y}_{1}}=c\] Chord \[x=9\] meets \[{{x}^{2}}-{{y}^{2}}=9\] at \[(9,\,\,6\sqrt{2})\] and \[(9,\,\,-6\sqrt{2})\] at which tangents are \[9x-6\sqrt{2}y=9\] and \[9x+6\sqrt{2}y=9\] or \[3x-2\sqrt{2}y-3=0\] and \[3x+2\sqrt{2}y-3=0\] \[\therefore \] Combined equation of tangents is \[(3x-2\sqrt{2}y-3)(3x+2\sqrt{2}y-3)=0\] or \[9{{x}^{2}}-8{{y}^{2}}-18x+9=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner