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I was going through Cut-the-Knot's "Joachimsthal's Notations" page discussing Joachimsthal's indice method of studying conics.

The conic can be represented in a following way:

$$ s_{ij} = Ax_i x_j + B(x_i y_j + x_j y_i) + C y_i y_j + F(x_i + x_j) + G(y_i + y_j) + H=0$$

Where the $i$ denotes a pair of number $(x_i,y_i)$ and $j$ also as $(x_j, y_j)$

It is a theorem that if $(x_i,y_i)$ is a point which lies on the conic, then ,

$$s_i = Ax_i x + B (x_i y + x y_i) + Cy_i y + F(x_i + x) + G(y_i + y) +H=0 \tag{1}$$

Where I have let the pair of $(x_j,y_j)$ to be set as free variable

Denotes the equation of tangent of the conic $$ s= Ax^2 +2Bxy + Cy^2 + 2Fx + 2Gy +H$$

At the point $(x_i,y_i)$

My question is regarding a step in the proof of this,

enter image description here

I don't get what exactly is the quantity benig plugged into (1) is, it seem the point has arguement $P(x_1,x_1)$ which doesn't make sense to me. I think the correct expression should be :

$$ P(x,y)= tP(x_1,y_1) + (1-t)P(x_2,y_2)$$

Hence the point being plugged in is given as:

$$ Q = (tx_1 + (1-t)x_2 , ty_1 + (1-t)y_2) $$

However, this seems to be still algebra bashing to move from (10) to (11), is there a more elegant approach to it?

tryst with freedom
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1 Answers1

5

This isn't particularly elegant, but it helps bring some order to the chaos of the symbol-bashing.

Introduce $u:=1-t$ to add a touch of symmetry to the linear combination of points: $$Q = t P(x_1,y_1) + u P(x_2,y_2) = \left(t x_1 + u x_2, t y_1 + u y_2\right) \tag1$$ Also, rewrite the quadratic expression by throwing-in convenient aliases of $1$: $$s = A x^2 + 2 B x y + C y^2 + 2 F (t+u) x + 2 G (t+u)y + H(t+u)^2 \tag2$$ Substituting $Q$ into this form of $s$ gives an expression we can write as $$\begin{array}{rll} A&(\color{red}{t x_1}+\color{blue}{u x_2})&(\color{red}{t x_1}+\color{blue}{u x_2}) \\ +2B&(\color{red}{t x_1}+\color{blue}{u x_2})&(\color{red}{t y_1}+\color{blue}{uy_2}) \\ +\phantom{2}C&(\color{red}{t y_1}+\color{blue}{u y_2})&(\color{red}{t y_1}+\color{blue}{u y_2}) \\ +2F&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)&(\color{red}{tx_1}+\color{blue}{u x_2}) \\ +2G&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)&(\color{red}{ty_1}+\color{blue}{uy_2}) \\ +\;H&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;) \end{array} \tag{3}$$ with which it's "easy" to gather terms by $\color{red}{t^2}$, $\color{blue}{u^2}$, and $\color{red}{t}\color{blue}{u}$ (without even explicitly expanding everything) to find $$t^2 s_{11} + u^2s_{22} + 2 tu\, s_{12} \tag4$$ Then, Cut-the-Knot's equation $(11)$ is just the above $(4)$ after recalling that $u=1-t$. $\square$

Note. One can re-skin expression $(3)$ using the matrix form of the conic equation $$X^TMX+NX+H$$ where $$M:=\begin{bmatrix}A& B\\ B&C\end{bmatrix}\qquad N:=\begin{bmatrix}2F& 2G\end{bmatrix}\qquad X:=\begin{bmatrix}x\\ y\end{bmatrix}$$ Substituting-in $X=tP_1+uP_2$ (where $P_i:=(x_i,y_i)$) gives $$(\color{red}{tP_1^T}+\color{blue}{uP_2^T})M(\color{red}{tP_1}+\color{blue}{uP_2})\;+\;N(\color{red}{t}+\color{blue}{u})(\color{red}{tP_1}+\color{blue}{uP_2})\;+\;H(\color{red}{t}+\color{blue}{u})(\color{red}{t}+\color{blue}{u}) \tag{3'}$$ where, again, it's "easy" to gather terms and get to $(4)$.

This still isn't particularly elegant, but it's a little more compact.

Blue
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