JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    If \[(\sqrt{3})bx+ay=2ab\]touches the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\]then the eccentric angle \[\theta \] of the point of contact is

    A)  \[\pi /6\]                            

    B)  \[\pi /2\]

    C)  \[3\pi /4\]                         

    D)  \[2\pi /3\]

    Correct Answer: A

    Solution :

     Idea Equation of tangent of an ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] at\[\theta \]i.e., \[(a\cos \theta ,b\sin \theta )\]is\[\frac{x}{a}\cos \theta +\frac{y}{b}\cos \theta =1\]Now, compare the both equations of tangent to get \[\theta \]. Here, equation of tangent \[\frac{x}{a}\frac{\sqrt{3}}{2}+\frac{y}{b}\frac{1}{2}=1\] and equation of tangent at the point (a\[\cos \theta ,b\sin \theta \]) is\[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] Both are same i. e., \[\cos \theta +\frac{\sqrt{3}}{2},\sin \theta =\frac{1}{2}\Rightarrow \]\[\theta =\frac{\pi }{6}\] TEST Edge Equation of normal, latus rectum, based questions are asked. To solve these types of questions, students are advised to understand the concept of an ellipse.

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