How to find type of conic section from its general equation?


by truongson243
Tags: conic, equation, type
truongson243
truongson243 is offline
#1
Feb23-12, 09:12 PM
P: 3
Given the general equation of conic section:
Ax2+Bxy+Cy2+Dx+Ey+F=0
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'2+C'y'2+D'x+E'y+F'=0
The new coefficients and parameters would be found from origin:
tan2θ=B/(A-C)
A', C' is the roots of quadratic equation: X2-(A+C)X+(4AC-B2)/4=0
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: x2+2xy+y2-8x+8y=0
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! :)
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Mark44
Mark44 is online now
#2
Feb24-12, 12:39 AM
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P: 20,967
Quote Quote by truongson243 View Post
Given the general equation of conic section:
Ax2+Bxy+Cy2+Dx+Ey+F=0
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'2+C'y'2+D'x+E'y+F'=0
The new coefficients and parameters would be found from origin:
tan2θ=B/(A-C)
A', C' is the roots of quadratic equation: X2-(A+C)X+(4AC-B2)/4=0
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: x2+2xy+y2-8x+8y=0
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.
This isn't a parabola - it's a circle. If A = C, the conic section is a circle.

As you already know, the xy term indicates a rotation. Let me assume that B = 0 so that there is no rotation.

If A = 0 and C [itex]\neq[/itex] 0, the conic is a parabola that opens left or right.
If C = 0 and A [itex]\neq[/itex] 0, the conic is a parabola that opens up or down.
Opening left vs right or up vs. down are determined by the agreement or disagreement in sign between the squared term and the linear term in the other variable.

For example, y2 + x = 0 (A = 0, C = 1, D = 1) is a parabola that opens to the left. For another example, x2 - y = 0 (A = 1, C = 0, E = 1) is a parabola that opens up.

If A and C are opposite in sign, the conic is a hyperbola.
If A and C are the same sign, but not equal, the conic is an ellipse.

If A = 0 and C = 0, the "conic" is a degenerate hyperbola (actually two straight lines).
Quote Quote by truongson243 View Post
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! :)
truongson243
truongson243 is offline
#3
Feb24-12, 01:32 AM
P: 3
Quote Quote by Mark44 View Post
This isn't a parabola - it's a circle. If A = C, the conic section is a circle.
Thank you so much but it's only a circle in case the term of xy is equal to zero. In this case, when I change the coordinate into which Ox' is incline 45 to Ox (horizontal), the new coefficients of general equation in new coordinate are A', C' (B' = 0 in the new coor. so that there is no rotation) satisfy: X2-2X=0, so I have to choose A' = 0 or C' = 0 and the parabola will open left or downward respectively. But, the computer only illustrated the case C=0, so we have the graph of function: y=-x2/4[itex]\sqrt{2}[/itex] (in new coordinate)
Why didn't it choose A'=0 so we'll have the graph of x=-y2/4[itex]\sqrt{2}[/itex], the open-left parabola or doesn't it exist?
Could you have a look at this link below? It's what my computer illutrated for the relation x2+2xy+y2-8x+8y=0 or, (x+y)2=8(x-y)

http://s880.photobucket.com/albums/a...t=parabola.jpg

Norwegian
Norwegian is offline
#4
Feb24-12, 02:53 AM
P: 144

How to find type of conic section from its general equation?


The type of the conic section can be read from its discriminant D=b2-4ac.

D>0: hyperbola
D=0: parabola
D<0: ellipse, incl circle as a special case when b=0, a=c.

This assumes the conic section exists over R, is irreducible.
(edit: and given by ax2+bxy+cy2+dx+ey+f=0)


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