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Math Help - Intersection of two surfaces

  1. #1
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    Intersection of two surfaces

    Note: This is from a Vector Calculus course, so I hope it belongs in this section.

    The question:
    For what values of 'a' do the surfaces x^2 + y^2 + (z - 1)^2 = 1 and a(z+1)^2 = x^2 + y^2, (z \ge -1) not intersect?

    My attempt:
    I rearranged the first equation like so:
    x^2 + y^2 = 1 - (z - 1)^2

    Then equated it with the other:
    a(z+1)^2 = 1 - (z - 1)^2

    Usually I'd solve for z to find the conditions for the intersection, but we have two variables now.

    I then solved for a:
    a = \frac{1 - (z - 1)^2}{(z+1)^2}

    So when z = -1, 'a' isn't defined. I then substituted this value into the very first equation to see what happens, but I just get the origin point.

    I don't think I'm doing this correctly. Any assistance would be great!
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  2. #2
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    a(z+1)^2 = 1 - (z - 1)^2

    (a+1)z^2+2(a-1)z+a=0

    This equation has no solution when the discriminant is negative
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  3. #3
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    Ahh that makes sense, thank you!
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  4. #4
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    Quote Originally Posted by Glitch View Post
    Note: This is from a Vector Calculus course, so I hope it belongs in this section.

    The question:
    For what values of 'a' do the surfaces x^2 + y^2 + (z - 1)^2 = 1 and a(z+1)^2 = x^2 + y^2, (z \ge -1) not intersect?

    My attempt:
    I rearranged the first equation like so:
    x^2 + y^2 = 1 - (z - 1)^2

    Then equated it with the other:
    a(z+1)^2 = 1 - (z - 1)^2

    Usually I'd solve for z to find the conditions for the intersection, but we have two variables now.

    I then solved for a:
    a = \frac{1 - (z - 1)^2}{(z+1)^2}
    This is the "wrong way". Solve the equation for z, not a. That is a quadratic equation for z so there will be no real number solution for z if the "discriminant" ( b^2- 4ac) is negative.

    So when z = -1, 'a' isn't defined. I then substituted this value into the very first equation to see what happens, but I just get the origin point.

    I don't think I'm doing this correctly. Any assistance would be great!
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