JEE Main & Advanced Mathematics Conic Sections Question Bank Hyperbola

  • question_answer
    If the tangent on the point \[(2\sec \varphi ,\ 3\tan \varphi )\] of the hyperbola \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{9}=1\] is parallel to \[3x-y+4=0\], then the value of f is [MP PET 1998]

    A)            \[{{45}^{o}}\]                            

    B)            \[{{60}^{o}}\]

    C)            \[{{30}^{o}}\]                            

    D)            \[{{75}^{o}}\]

    Correct Answer: C

    Solution :

               Differentiation of \[x=2\sec \varphi \] Þ\[\frac{dx}{d\varphi }=2\sec \varphi \tan \varphi \] Differentiate, \[y=3\tan \varphi \]w.r.t. f, we get\[\frac{dy}{d\varphi }=3{{\sec }^{2}}\varphi \] \[\therefore \]Gradient of tangent \[\frac{dy}{dx}=\frac{dy/d\varphi }{dx/d\varphi }=\frac{3{{\sec }^{2}}\varphi }{2\sec \varphi \tan \varphi }\] \[\frac{dy}{dx}=\frac{3}{2}\,\text{cosec}\varphi \]                                                           .....(i) But, tangent is parallel to \[3x-y+4=0\] \[\therefore \]Gradient \[m=3\]                                              .....(ii) By (i) and (ii), \[\frac{3}{2}\text{cosec}\varphi =3\]Þ\[\text{cosec}\varphi =2\], \[\therefore \varphi =30{}^\circ \].


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