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anonymous

  • 5 years ago

If A be a point on the ellipse x^2/a^2 + y^2/b^2 = 1 and B be a point on its auxillary circle vertically above it, then find the locus of the midpt of A,B

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  1. anonymous
    • 5 years ago
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    The equation for the auxiliary circle is x^2+y^2=a^2

  2. anonymous
    • 5 years ago
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    i want that locus even i know the aux circle

  3. anonymous
    • 5 years ago
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    Don't be smart-a## to someone trying to help.

  4. anonymous
    • 5 years ago
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    k sorry...cmon..

  5. anonymous
    • 5 years ago
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    Some how related to Kepler's laws of planetary motion. Find the radius of the ellipse, radius of the aux: they form two legs of a right triangle. The distance between A, B is hypotenuse.

  6. anonymous
    • 5 years ago
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    im getting that but m still nowhere close 2 getting the locus..thnx anyway..

  7. anonymous
    • 5 years ago
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    I find that a=b and the distance AB is\[\sqrt{a ^{2}+b ^{2}}\]If you draw the ellipse on an x-y coordinate, you find that x is length a and y is length b, similarly the aux is equal to a. Once again a and b are legs of the right triangle.

  8. anonymous
    • 5 years ago
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    the locus is - \[y^2(a^2-b^2) = 0\] if a point lies on two curves then the locus of the point is the elimination of the equations of the curves. You have \[x^2b^2 + y^2a^2 = a^2b^2\] and \[x^2+y^2=a^2\] solve them up!

  9. anonymous
    • 5 years ago
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    no uve got it wrong man..i need the locus of the midpnt of the two points

  10. anonymous
    • 5 years ago
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    ahh.. i realise my mistake. I'll try again and let you know

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