JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer If the line \[y=mx+7\sqrt{3}\]is normal to the hyperbola \[\frac{{{x}^{2}}}{24}-\frac{{{y}^{2}}}{18}=1,\]then a value of m is [JEE Main 9-4-2019 Morning]

    A) \[\frac{\sqrt{5}}{2}\]                         

    B) \[\frac{3}{\sqrt{5}}\]

    C) \[\frac{2}{\sqrt{5}}\]             

    D) \[\frac{\sqrt{15}}{2}\]

    Correct Answer: C

    Solution :

    \[\frac{{{x}^{2}}}{24}-\frac{{{y}^{2}}}{18}=1\Rightarrow a=\sqrt{24};b=\sqrt{18}\] Parametric normal : \[\sqrt{24}\cos \theta .x+\sqrt{18}.y\,cot\theta =42\] At\[x=0:y=\frac{42}{\sqrt{18}}\tan \theta =7\sqrt{3}\](from given equation) \[\Rightarrow \tan \theta =\sqrt{\frac{3}{2}}\Rightarrow \sin \theta =\pm \sqrt{\frac{3}{5}}\] slope of parametric normal\[=\frac{-\sqrt{24}\cos \theta }{\sqrt{18}\cot \theta }=m\] \[\Rightarrow m=-\sqrt{\frac{4}{3}}\sin \theta =-\frac{2}{\sqrt{5}}or\frac{2}{\sqrt{5}}\]          

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