J & K CET Engineering J and K - CET Engineering Solved Paper-2008

  • question_answer
    A point P moves so that the sum of its distances from \[(-ae,\,0)\]and \[(ae,\,0)\] is\[2a\]. Then the locus of P is          

    A)  \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{x}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\]

    B)  \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\]

    C)  \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}(1+{{e}^{2}})}=1\]

    D)  \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{a}^{2}}(1+{{e}^{2}})}=1\]

    Correct Answer: B

    Solution :

    By using the properties, the sum of focal distance of any points on the ellipse is equal to the major axis. \[\therefore \]  Required equation is \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=1\]


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