Math Questions -- Algebra

Discussion in 'Physics & Math' started by Amadeusboy, Apr 4, 2008.

  1. Amadeusboy Registered Member

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    22
    Temur,
    Thanks for taking the time to respond. Knowing positively that \(p\) must be a constant would have simplified the question. I should have recognized this from the start. Oh well, live and learn.

    Thanks again,

    Amadeusboy
     
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  3. Amadeusboy Registered Member

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    I have a quick question. What would be the graph of

    \(\mid y+2\mid = \mid x-3\mid\)?

    Is is an "X" centered at \((-2,3)\) and having a 45 degree angle with \(y=-2\)?

    Thanks in advance for any help.

    Amadeusboy
     
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  5. temur man of no words Registered Senior Member

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    Yes. I think there is a typo: it is X centered at (3,-2).
     
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  7. Amadeusboy Registered Member

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    Thanks, Temur. You're right; there is a typo. It should be centered at \((3,-2)\).

    Amadeusboy
     
  8. Amadeusboy Registered Member

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    22
    I came across another couple questions with which I had problems. Again, they are taken from p. 50 of Boyce and DiPrima Calculus (1988). Note that I added the bracketed definitions.

    First,
    Find an equation satisfied by all points twice as far from \((-1,3)[=A]\) as from \((4,2)[=B]\).​

    and secondly,

    Find an equation satisfied by all points equidistant from the origin and the line \(x=-2\).​

    In the first problem, I think my difficulty lies in the interpretation . Clearly, the equation will be some sort of circle, but is it centered on one of the two points mentioned or should one randomly select a point \(P=(x,y)\) and find \(x\) and \(y\) such that \(d(P,A)=2d(P,B)\). (I haven't done the algebra, but this will be some sort of circle.)

    For the second problem, the graph of the equation in question must lie to the right of the line \(x=-2\). Thus, any point satisfying this equation will have a distance of \(x+2\) from the line \(x=-2\) and a distance \(\sqrt{x^2+y^2}\) from the origin. Would setting these two distances equal to each other and solving give the desired result? Is \(x=y^2/4+-1\) the correct answer?

    Thanks in advance for any assistance.

    Amadeusboy
     
  9. temur man of no words Registered Senior Member

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    1,330
    The latter version is correct. Finding an equation for a curve means finding a relation between x and y where (x,y) is a point on the curve. (I doubt it is a circle though)

    Yes, correct.
     
  10. D H Some other guy Valued Senior Member

    Messages:
    2,257
    I suggest that you work with the square of the distances rather than distances because that way you won't have those nasty radicals to deal with. With this, your equation becomes \(d^2(P,A)=4d^2(P,B)\) or \(4d^2(P,B) - d^2(P,A) = 0\), where \(d^2(P,A) = (x+1)^2+(y-3)^2\) and similar for \(d^2(P,B)\). Note that \(d^2[tex] is a positive definite form. In other words, you do not have to worry about a negative values for [tex]d^2\) (and worrying about them well get you in trouble, like it did on the next problem.)

    No. That +- doesn't belong. Look at it this way. Set y to 0. There is only one point on the x-axis that is equidistant from the origin and the line x=-2: The point (-1,0). Your equation says (1,0) is also a solution. You got in trouble by worrying about radicals having two solutions when you should not have done so.

    The answer to the first problem is a circle and the second, a parabola. In fact, the second answer is the standard geometric definition of a parabola as the locus of all points that are equidistant between a line and a point. The line is called the directrix of the parabola and the point, the focus of the parabola.
     
  11. Amadeusboy Registered Member

    Messages:
    22
    Temur and D H, thanks for responding.

    D H, the "+" sign in my final solution to the second problem is yet another typographical error. Thank you reminding me of the geometric definition of a parabola. I had forgotten this (along with a great deal of other things).

    Thanks again,
    Amadeusboy
     

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