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Given the general equation of conic section:

Ax

How can we find the type of conic section from the above equation?

I've used method of rotating coodinate around the origin (from Oxy to Ox'y', Ox' makes an angle θ to Ox, counterclockwise) such that the symmetry axes of the conic curve are Ox' and Oy', respectively.

The general equation of conic section in the new coodinate without coefficient B':

A'x'

The new coefficients and parameters would be found from origin:

tan2θ=B/(A-C)

A', C' is the roots of quadratic equation: X

D'=Dcosθ+Esinθ

E'=-Dsinθ+Ecosθ

F'=F.

And then find the type of conic section normally.

But when I applied this method, there was something strange.

Consinder the relation: x

From above, we can find: θ=π/4 or 3π/4 (There're two orientations of Ox'y')

A', C' are found from: X(X-2)=0. So A', C' can be alternative. The parabola has a paralleil to Ox' or Oy' is symmetric axis depends on which value of A' and C' we choose.

So, we can find four cases of orientation of this parabola. But the graph I obtained using graph draw software just only gave one case. (The vertex of parabola direct toward the quad-corner II(upper left corner))

I do not understand at all. Could you explain me why does it ignore the other cases and do they actually exist?

Thank you so much! :)

Ax

^{2}+Bxy+Cy^{2}+Dx+Ey+F=0How can we find the type of conic section from the above equation?

I've used method of rotating coodinate around the origin (from Oxy to Ox'y', Ox' makes an angle θ to Ox, counterclockwise) such that the symmetry axes of the conic curve are Ox' and Oy', respectively.

The general equation of conic section in the new coodinate without coefficient B':

A'x'

^{2}+C'y'^{2}+D'x+E'y+F'=0The new coefficients and parameters would be found from origin:

tan2θ=B/(A-C)

A', C' is the roots of quadratic equation: X

^{2}-(A+C)X+(4AC-B^{2})/4=0D'=Dcosθ+Esinθ

E'=-Dsinθ+Ecosθ

F'=F.

And then find the type of conic section normally.

But when I applied this method, there was something strange.

Consinder the relation: x

^{2}+2xy+y^{2}-8x+8y=0From above, we can find: θ=π/4 or 3π/4 (There're two orientations of Ox'y')

A', C' are found from: X(X-2)=0. So A', C' can be alternative. The parabola has a paralleil to Ox' or Oy' is symmetric axis depends on which value of A' and C' we choose.

So, we can find four cases of orientation of this parabola. But the graph I obtained using graph draw software just only gave one case. (The vertex of parabola direct toward the quad-corner II(upper left corner))

I do not understand at all. Could you explain me why does it ignore the other cases and do they actually exist?

Thank you so much! :)

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