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Math Help - Finding the center of conic sections

  1. #1
    Junior Member Dergyll's Avatar
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    Finding the center of conic sections

    Hey guys, just ran across this question on a homework:

    "Find the center of the conic section whose equation is: A x +B xy +C y +D x +E y +F=0"

    How do I find them without any numbers in each term? Is there a general formula to solve for the center like -b/2a?

    Any help is appreciated
    Derg
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Dergyll View Post
    Hey guys, just ran across this question on a homework:

    "Find the center of the conic section whose equation is: A x +B xy +C y +D x +E y +F=0"

    How do I find them without any numbers in each term? Is there a general formula to solve for the center like -b/2a?

    Any help is appreciated
    Derg
    This is messy but you can complete the square
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dergyll View Post
    Hey guys, just ran across this question on a homework:

    "Find the center of the conic section whose equation is: A x +B xy +C y +D x +E y +F=0"

    How do I find them without any numbers in each term? Is there a general formula to solve for the center like -b/2a?

    Any help is appreciated
    Derg
    Define what you mean by "center." For example I have never heard of the term used for a parabola. (Though there is a point called the focus that might serve a similar function.)

    If B = 0 then it's fairly easy. As Mathstud28 suggested you can complete the square on the x and/or y terms (as applicable) and simplify from there. If B is not 0 then you need to rotate the coordinate system such that the new form has the x'y' coefficient equal to 0.

    The question is odd. There is no way to approach this until at least some of the coefficients are known.

    -Dan
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  4. #4
    Junior Member Dergyll's Avatar
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    That is exactly word for word what the question says. The center (because of the fact that the graph is NOT a parabola since both x and y are both squared) should be that of an eclipse. But I still don't know how to solve for it!!! Does the problem want me to plug numbers in?

    Thanks
    Derg
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  5. #5
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    Quote Originally Posted by Dergyll View Post
    [snip]
    The center (because of the fact that the graph is NOT a parabola since both x and y are both squared) should be that of an eclipse. But I still don't know how to solve for it!!!
    [snip]
    It makes sense to talk about the centre of a hyperbola too, you know - it's the intersection point of its asymptotes .......

    Quote Originally Posted by Dergyll View Post
    [snip]
    Does the problem want me to plug numbers in?

    Thanks
    Derg
    Not being able to consult with the author of the question, who knows? Although, if that's what the question wanted you to do, shouldn't it say so ....?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dergyll View Post
    That is exactly word for word what the question says. The center (because of the fact that the graph is NOT a parabola since both x and y are both squared) should be that of an eclipse. But I still don't know how to solve for it!!! Does the problem want me to plug numbers in?

    Thanks
    Derg
    What graph? And the equation you posted was in terms of unknown constants A, B, ... You didn't state any conditions on them so some of them may be zero and may be positive or negative. The equation you gave was the general form for a conic section: it could be anything.

    -Dan
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  7. #7
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    Hello, Derg!

    If you haven't been taught about Rotations,
    . . this problem is inappropriate.


    Find the center of the conic section whose equation is:
    . . . Ax^2 + Bxy + Cy^2 + DX + Ey + F \;=\;0
    There is a "discriminant": . \Delta \;=\;B^2-4AC\qquad \begin{Bmatrix}\Delta = 0 & \text{Parabola} \\ \Delta > 0 & \text{Hyperbola} \\ \Delta < 0 & \text{Ellipse} \end{Bmatrix}

    If there is no xy-term (B = 0), it is a standard conic.
    . . Its axes are parallel to the coordinate axes.


    If B \neq 0, the conic has been rotated through angle \theta

    . . where: . \tan2\theta \:=\:\frac{B}{A-C}


    We can "un-rotate" the graph with these conversions:

    . . \begin{array}{ccc} X &=& x\cos\theta - y\sin\theta \\ Y &=& x\sin\theta - y\cos\theta \end{array}


    We also have:

    . . A' \;=\;A\cos^2\!\theta + B\sin\theta\cos\theta + C\sin^2\!\theta
    . . B' \;=\;0
    . . C' \;=\;A\sin^2\!\theta - B\sin\theta\cos\theta + C\cos^2\!\theta
    . . D' \;=\;D\cos\theta + E\sin\theta
    . . E' \;=\;\text{-}D\sin\theta + E\cos\theta
    . . F' \;=\;F


    We get a new equation: . A'X^2 + C'Y^2 + D'X + E'Y + F' \;=\;0

    . . and now you can complete the square, etc.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Note: They always use the term "center".
    . . . . .In a parabola, it refers to the vertex.

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