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    The question is:

    Given z=x+iy satisfies |z-4| + |z+4| = 10, show that (\frac{x}{5})^2 + (\frac{y}{3})^2 = 1.
    I've got a complete mind blank at the moment. :sigh:
    I know that

    |z-4| = \sqrt{(x-4)^2 + y^2}\ \ \ \ \mathrm{and}\ \ \ \ |z+4| = \sqrt{(x+4)^2 + y^2},

    but how would I combine the two square roots to get something like what I'm supposed to show?
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    (Original post by james.h)
    how would I combine the two square roots to get something like what I'm supposed to show?
    Do what you think you are supposed to do. Square stuff and plod through the algebra. I cheated by using a computer but can confirm that it all simplifies nicely to 36x^2 + 100y^2 = 900 which is the ellipse given.
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    (Original post by Mr M)
    Do what you think you are supposed to do. Square stuff and plod through the algebra. I cheated by using a computer but can confirm that it all simplifies nicely to 36x^2 + 100y^2 = 900 which is the ellipse given.
    :yep: I've got it now. Thanks.

    That question was horrible. :sad:
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    (Original post by james.h)
    :yep: I've got it now. Thanks.

    That question was horrible. :sad:
    Good.
 
 
 
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