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

In summary, the type of conic section can be determined by finding the discriminant of the general equation, where D>0 indicates a hyperbola, D=0 indicates a parabola, and D<0 indicates an ellipse or circle as a special case. However, this assumes the conic section exists over R, is irreducible, and is given by the general equation ax2+bxy+cy2+dx+ey+f=0.
  • #1
truongson243
3
0
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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  • #2
truongson243 said:
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).
truongson243 said:
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! :)
 
  • #3
Mark44 said:
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/ac10/zarya_243/?action=view&current=parabola.jpg
 
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  • #4
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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  • #5


There are a few things to consider when using this method to find the type of conic section from its general equation. First, it is important to note that when we rotate the coordinate system, we are essentially changing the orientation of the conic section. This means that we may end up with different coefficients and parameters depending on the chosen angle of rotation.

In the specific example given, it is possible that the software you are using is only showing one orientation of the parabola. This could be due to the limitations of the software or the specific settings being used. It is also possible that there is a mistake in the calculation of the coefficients and parameters, leading to the discrepancy between the expected and observed orientations of the parabola.

Additionally, it is important to remember that the general equation of a conic section can represent different types of conic sections depending on the values of the coefficients. For example, a general equation with A and C equal to zero would represent a line, while A and C being unequal would represent a hyperbola or an ellipse. Therefore, it is important to consider the values of A and C when determining the type of conic section from its general equation.

In summary, while this method can be used to determine the type of conic section from its general equation, it is important to carefully consider the chosen angle of rotation and the values of the coefficients in order to accurately determine the type of conic section. It is also important to check for any potential mistakes in the calculations and to consider the limitations of the software being used.
 

1. What is a conic section?

A conic section is a shape that is created when a plane intersects with a cone. The resulting shape can be a circle, ellipse, parabola, or hyperbola, depending on the angle of the intersecting plane.

2. How can I identify the type of conic section from its general equation?

To identify the type of conic section from its general equation, you can look at the powers of the variables in the equation. A second degree equation (x^2 or y^2) indicates a circle, ellipse, or hyperbola, while a first degree equation (x or y) indicates a parabola.

3. What are the key features of a circle?

A circle has a center point, and all points on the circle are equidistant from the center. Its general equation is (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.

4. How can I determine the center and radius of a circle from its general equation?

To determine the center and radius of a circle from its general equation, you can rearrange the equation into the standard form, (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.

5. What is the focus-directrix property of a parabola?

The focus-directrix property of a parabola states that all points on a parabola are equidistant from the focus point and the directrix line. The general equation for a parabola is y = ax^2 + bx + c, where a determines the direction and shape of the parabola, and (h,k) is the vertex point which is also the focus point.

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