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JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    If the line \[y=ax+b\] is tangent to the curve \[f(x)=x-{{x}^{3}}\]at point \[(-1,0)\]then the eccentricity of the ellipse \[\frac{{{x}^{2}}}{{{a}^{4}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]  is

    A)  \[\frac{1}{2}\]                         

    B)  \[\frac{1}{\sqrt{2}}\]

    C)  \[\frac{\sqrt{3}}{2}\]                

    D) \[\frac{2}{3}\]

    Correct Answer: C

    Solution :

                \[\frac{dy}{dx}=1-3{{x}^{2}}\,{{|}_{(-1,\,0)}}=-2\] \[\therefore \,\,a=-2\] Also (-1, 0) lies on line \[0=b-a\] \[\therefore \,\,b=-2\] \[E:\frac{{{x}^{2}}}{16}\,+\frac{{{y}^{2}}}{4}=1\] \[{{e}^{2}}=1-\frac{4}{16}\,=1-\frac{1}{4}\,=\frac{3}{4}\] \[e\frac{\sqrt{3}}{2}\]


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