CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    If for the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1,\]\[y-\]axis is the minor axis and the length of the latus rectum is one half of the length of its minor axis, then its eccentricity is

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

    B)  \[\frac{1}{2}\]

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

    D)         \[\frac{3}{4}\]

    E)  \[\frac{3}{5}\]

    Correct Answer: C

    Solution :

    We know that length of latusrectum of an ellipse\[=\frac{2{{b}^{2}}}{a}\]and length of its minor axis\[=2b\] \[\therefore \]  \[\frac{2{{b}^{2}}}{a}=b\Rightarrow 2b=a\] Also,  \[e=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{{{b}^{2}}}{4{{b}^{2}}}}\]                 \[=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\]


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