The degree-2 curves through three points is given by $\gamma L_1L_2+\lambda L_2L_3+\mu L_3L_1=0$, where the $L_i$ are the lines joining those points

Question

I came across this equation while reading on conics and circles.

The family of degree-$2$ curves passing through three points is given by $$\gamma L_1L_2 + \lambda L_2L_3 + \mu L_3L_1 = 0$$ where $L_1$, $L_2$, and $L_3$ are the equations of lines joining the three points, and $\lambda$, $\mu$ and $\gamma$ are parameters.

I can see that the three points definitely satisfy this equation , but I am unable to say much about rest of the points.

How would you prove this relation? Explain, please. :-)

Answer 1

Conics can be expressed in terms of trilinear coordinates, which express points in terms of relative directed distances from three lines (which can be thought of as the sidelines of a triangle. In these terms it is much easier to reason about the relations between a family of conics and a given triangle. Variations of the formula given in the OP will, for example, give expressions for the conics that are tangential to a given triangle, or have that triangle as a self polar triangle.

A good reference for this was published over 150 years ago: Whitworth's Trilinear Coordinates. The URL points to Chapter XV, which discusses the case in the OP. Using the key $\alpha\rightarrow L_1, \beta\rightarrow L_2, \gamma\rightarrow L_3, l\rightarrow \lambda, m\rightarrow \mu, n\rightarrow \gamma,$ the equation $l\beta\gamma+m\gamma\alpha+n\alpha\beta=0$ translates directly into the OP equation.

But how do we get there from here, namely a conic (expressed in homogeneous coordinates) $$Ax^2 + By^2 + 2Gxz + 2Fyz + 2Hxy + Cz^2 = 0 \tag{1}$$ that may or may not go through the three points?

The answer is a change of variables. We know that (again, using homogeneous coordinates) $$ \begin{align} L_1 &= a_1 x+b_1 y + c_1z \\ L_2 &= a_2 x+b_2 y + c_2z \\ L_3 &= a_3 x+b_3 y + c_3z, \\ \end{align} $$

so we can solve for $x,y,z$ in terms of $L_1,L_2,L_3$ and then plug the result into (1) to get an equation of the form

$$ fL_1^2+gL_2^2+hL_3^2+2lL_2L_3+2mL_3L_1+2nL_1L_2=0,\tag{2}\ $$

which maps (using the key given earlier) to the second equation on pg 192. Assuming that the conic (1) passes through the three points we can use the reasoning on that page to determine that $f=g=h=0,$ which reduces (2) to

$$ lL_2L_3+mL_3L_1+nL_1L_2=0. $$

This answer could have been given without reference to Whitworth, but I think it's important to underline the use of trilinear coordinates. For a point $P=(x,y,z),L_n(P)$ gives a relative directed distance of point $P$ from line $L_n$, and can be thought of as a trilinear coordinate. The cases of conic families that inscribe the triangle, or for whom the triangle is self-polar are treated in Chapters XVI and XIV of the reference. Also note that trilinear coordinates are projective/homogeneous, so it's easier to make this argument using homogeneous coordinates $(x,y,z)$ rather than Cartesian coordinates $(x/z,y/z).$

Answer 2

Simple explanation for pairwise satisfying the assembly of straight lines and conic generation from lines .

We know that

$$ \lambda' L_2 + \mu' L_3 = 0$$

represents a straight line passing through the intersection point of $L_2,L_3.$

Divide given relation by $L_1L_2L_3$

$$\gamma/ L_3 + \lambda/ L_1 + \mu /L_2 = 0$$

Extending the same logic for a pair we have a single curve passing through all three intersection points with another combination parameter set.

Setting one parameter zero three pair-wise solutions for three sets of straight lines gives three points for the vertices of a triangle.

Given second degree equation is a conic which must satisfy/pass through these three points. That is the triangle is inscribable or escribable in the conic... in a sense as being concurrent.