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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    If the line \[x2y=12\]is tangent to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]at the point \[\left( 3,\frac{-9}{2} \right),\]then the length of the latus recturm of the ellipse is : [JEE Main 10-4-2019 Morning]

    A) \[9\]     

    B) \[8\sqrt{3}\]

    C) \[12\sqrt{2}\]              

    D) \[5\]

    Correct Answer: A

    Solution :

                Tangent at\[\left( 3,-\frac{9}{2} \right)\] \[\frac{3x}{{{a}^{2}}}-\frac{9y}{2{{b}^{2}}}=1\] Comparing this with \[x2y=12\] \[\frac{3}{{{a}^{2}}}=\frac{9}{4{{b}^{2}}}=\frac{1}{12}\] we get a = 6 and \[b=3\sqrt{3}\] \[L(LR)=\frac{2{{b}^{2}}}{a}=9\]          


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