Using the discriminants of conic sections

Question

According to this page , we can understand the shape of conic sections using the discriminants of general equation but how we can prove it ? I'm really confused about that complicated formulas without any proof .

Answer 1

If we have a general equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, and $B\neq 0$ we rotate by $\theta$ given by $\tan{2\theta}=\frac{B}{A-C}$to eliminate this $xy$-term: the discriminant $B^2-4AC$ is invariant as we can check directly by identifying new $A',B',C',D',E',F'$ through the transformation calculate $B'^2-4A'C'$.

Now for a conic with $B=0$ we have a parabola if $A=0$ or $C=0$, i.e. $AC=0$ or they are both non-zero and we have an ellipse or circle if $A$ and $C$ have the same sign i.e. $AC>0$, and finally we have a hyperbola if they have the opposite sign i.e. $AC<0$.

Transforming to an equation with $B'=0$ through rotation, we can see that

$$B^2-4AC=B'^2-4A'C'=-4A'C'$$ since we rotate to a situation where $B'=0$. Now we can read off the criteria:

parabola: $B^2-4AC=0$

ellipse or circle: $B^2-4AC<0$

hyperbola: $B^2-4AC>0$.

Answer 2

For variety, we can prove this with the geometry of the complex projective plane.

Nondegenerate conics given by quadratic equations with real coefficients can be distinguished by how they intersect the line at infinity. There are two points of intersection with multiplicity, and they can be of the following types:

• A hyperbola meets the line at infinity in two real points, corresponding to the two asymptotes.
• A parabola is tangent to the line at infinity, and so intersects the line at infinity at a single point with multiplicity 2
• An ellipse doesn't meet the line at infinity over the reals; the intersection consists of a complex conjugate pair of points. (An ellipse has asymptotes with complex slope)

If we take the equation for the general quadratic and homogenize it, we have the equation

$$ Ax^2+Bxy+Cy^2+Dxz+Eyz+Fz^2=0$$

The line at infinity is given by the equation $z=0$. Substituting into the equation gives

$$ Ax^2+Bxy+Cy^2 = 0$$

Solving this quadratic equation, we get:

• Two real solutions when the discriminant is positive — thus hyperbola
• A double solution when the discriminant is zero — thus parabola
• Two complex solutions when the discriminant is negative — thus ellipse