Question

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.

Answer 1

\[2x^2+2\sqrt(3)xy-y^2+4x-y = 10\]

Answer 2

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

Answer 3

rotate it back lol

Answer 4

uzma, it is actually \[H^2 - 4AB\]

Answer 5

its the same as the discriminant of a quadratic

Answer 6

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

Answer 7

oh, ok didn't see that well :/

Answer 8

A=2
B=-1
H=sqrt3
H^2-AB=3+2=5
so its ellipse

Answer 9

it's a hyperbola when it's positive :)

Answer 10

awwwww....u right:)

Answer 11

any one cares to find the angle of rotation needed to rotate it back ?

Answer 12

what does it actually mean?

Answer 13

ok my way of saying was vague

Answer 14

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 \[Ax^2 + By^2 +Dx+Ey +F = 0\]

Answer 15

becomes my conic where

x' = xcos(a) + ysin(a)
y' = -xsin(a) + ycos(a)

a is my angle of rotation

Answer 16

ah..rotation of axes :)

Answer 17

so u mean to transform the eq in the new system so that there isnt a mixed term?