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yuki

  • 5 years ago

Do you guys want to solve a fun problem? Find the angle of rotation that is needed for the following conic and find out the type.

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  1. Yuki
    • 5 years ago
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    \[2x^2+2\sqrt(3)xy-y^2+4x-y = 10\]

  2. anonymous
    • 5 years ago
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    the type of conic is determined by the sign of H^2-AB, where H=coefficient of xy A=coeff of x^2 B=coeff of y^2

  3. amistre64
    • 5 years ago
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    rotate it back lol

  4. Yuki
    • 5 years ago
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    uzma, it is actually \[H^2 - 4AB\]

  5. Yuki
    • 5 years ago
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    its the same as the discriminant of a quadratic

  6. anonymous
    • 5 years ago
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    it is also H^2-AB the difference is that in ur case H is coefficient of xy in the second case it is the coefficient of 2xy

  7. Yuki
    • 5 years ago
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    oh, ok didn't see that well :/

  8. anonymous
    • 5 years ago
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    A=2 B=-1 H=sqrt3 H^2-AB=3+2=5 so its ellipse

  9. Yuki
    • 5 years ago
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    it's a hyperbola when it's positive :)

  10. anonymous
    • 5 years ago
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    awwwww....u right:)

  11. Yuki
    • 5 years ago
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    any one cares to find the angle of rotation needed to rotate it back ?

  12. anonymous
    • 5 years ago
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    what does it actually mean?

  13. Yuki
    • 5 years ago
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    ok my way of saying was vague

  14. Yuki
    • 5 years ago
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    whenever the xy term has a coefficient other than 0 you can rotate the xy-coordinate system to make the equation have no xy term so that \[Ax^2 + By^2 +Dx+Ey +F = 0\]

  15. Yuki
    • 5 years ago
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    becomes my conic where x' = xcos(a) + ysin(a) y' = -xsin(a) + ycos(a) a is my angle of rotation

  16. anonymous
    • 5 years ago
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    ah..rotation of axes :)

  17. anonymous
    • 5 years ago
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    so u mean to transform the eq in the new system so that there isnt a mixed term?

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