2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Conic Sections

JEE Main & Advanced Mathematics Conic Sections Question Bank Self Evaluation Test - Conic Sections

  • question_answer
    If \[P(\theta )\] and \[Q\left( \frac{\pi }{2}+\theta  \right)\] are two points on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], then locus of the mid-point of PQ is

    A) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=\frac{1}{2},\]

    B) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=4\]

    C) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=2\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] Clearly P is \[(acos\theta ,bsin\theta )\]and Q is \[(-asin\theta ,b\sin \theta )\] so the midpoint (h, k) or PQ will be given by \[h=\frac{a\cos \theta -a\sin \theta }{2}\] And \[k=\frac{b\sin \theta +b\cos \theta }{2}\] \[\therefore \frac{4{{h}^{2}}}{{{a}^{2}}}+\frac{4{{k}^{2}}}{{{b}^{2}}}=2\Rightarrow \frac{{{h}^{2}}}{{{a}^{2}}}+\frac{{{k}^{2}}}{{{b}^{2}}}=\frac{1}{2}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner