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Find the equation of the following. The locus of all points in a plane such that the sum of the distances from points (-2,-1) and (-10,-1) is equal to 6.

  • calculus -

    I think you have an impossible situation
    the distance between the two given points is 8 units

    The sum of the distance from any point to the two given points has to be greater than 8

  • calculus -

    Oh shoot. You're right. I took the last part from a different problem. The sum of the distances is equal to 10. Sorry.

  • calculus -

    Now it becomes an ellipse.
    from the data
    2a = 10
    a = 5
    the two given points are the foci
    and the centre would have to be the midpoint of those, namely
    the distance form the centre to one focal point is the c value, here it would be 4
    In an ellipse with the a horizontal major axis
    b^2 + c^2 = a^2
    b^2 + 16 = 25
    b = 3

    in standard form then
    (x+6)^2/25 + (y+1)^2/9 = 1

  • calculus -

    Thanks, but where did you get the

    2a = 10


  • calculus -

    by definition, the sum of the two focal length is 2a

    check, distance from centre to the x vertex is a + a = 2a

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