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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    If the tangents on the ellipse \[4{{x}^{2}}+{{y}^{2}}=8\]at the points (1, 2) and (a, b) are perpendicular to each other, then \[{{a}^{2}}\] is equal to :               [JEE Main 8-4-2019 Morning]

    A) \[\frac{64}{17}\]                                

    B) \[\frac{2}{17}\]

    C) \[\frac{128}{17}\]                              

    D) \[\frac{4}{17}\]

    Correct Answer: B

    Solution :

    \[4{{a}^{2}}+{{b}^{2}}=8\]                      ....(a) also\[{{\left. \frac{dy}{dx} \right)}_{(1,2)}}=-\frac{4x}{y}=-2\] \[\Rightarrow \]\[-\frac{4a}{b}=\frac{1}{2}\] \[b=-8a\] \[\Rightarrow \]\[{{b}^{2}}=64{{a}^{2}}\] \[68{{a}^{2}}=8\] \[{{a}^{2}}=\frac{2}{17}\]


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