Hyperbola locus sketched by Descartes

Question

D.J.Struik's A Concise History of Mathematics (pp 98 4th edition) contains a sketch I modified here (wrongly?) Variable line $GPL$ passes through a fixed point $G$ and, as a rigid right triangle $LNK$ moves on x-axis, $P$ traces out a hyperbola with Cartesian equation:

$$ y^2 = cy - (c/b) \,x \,y + a y -ac $$

Did I get it right?

EDIT1:

Original French was long winding, Struik's translation elusive (to me) with a print error.Today it goes so fast, so posted to check it out.. an oblique asymptoted hyperbola, axes tilted.

EDIT2:

Subsequent to this post hand sketched 3 types of tiny right triangles.

Answer 1

Let's see.

\begin{align*} G &= \begin{pmatrix}0\\a\end{pmatrix} & L &= \begin{pmatrix}t\\0\end{pmatrix} & K &= \begin{pmatrix}t+b\\0\end{pmatrix} & N &= \begin{pmatrix}t\\c\end{pmatrix} \end{align*}

\begin{alignat*}{2} GL&:\quad & ax + ty &= at \\ KN&:\quad & cx + by &= c(t+b) \\ \end{alignat*}

\begin{equation*} P = \frac1{ab-tc}\begin{pmatrix} t(ab-tc-bc)\\abc \end{pmatrix} \end{equation*}

Assuming $a,b,c\neq 0$ this means the locus of $C$ has to be

$$abc - aby - bcy + cxy + by^2 = 0$$

which is equivalent to the formula you gave. So yes, you are right.

How did I compute this? I represented my points as homogeneous coordinates, e.g. $G=[0:a:1]$. Then computing the line joining two points means computing the cross product. intersecting two lines is a cross product, too. So at that point I had $P=[P_x:P_y:P_z]$ in homogeneous coordinates. Usually one would dehomogenize this using $x=\frac{P_x}{P_z}$ but instead I kept this as a polynomial, namely $P_zx - P_x=0$, and likewise $P_zy - P_y=0$. Then I eliminated $t$ from these two polynomials using a resultant.

\begin{align*} P_x &= t(ab-tc-bc) \\ P_y &= abc \\ P_z &= ab-tc \\ P_zx - P_x &= c t^{2} + \left(- a b + b c - c x\right) t + a b x \\ P_zy - P_y &= - c y t - a b c + a b y \\ 0 &= \begin{vmatrix} c & - a b + b c - c x & a b x \\ - c y & - a b c + a b y & 0 \\ 0 & - c y & - a b c + a b y \end{vmatrix} \\ &= b \cdot a \cdot c^{2} \cdot (a b c - a b y - b c y + c x y + b y^{2}) \end{align*}

Answer 2

\begin{align} \triangle PXK \sim \triangle NLK &\implies \dfrac{y}{XK} = \dfrac cb \\ &\implies XK = \dfrac bcy \\ &\implies XL = \dfrac bcy-b \\ \end{align}.

\begin{align} \triangle GYP \sim \triangle PXL &\implies \dfrac{a-y}{x} = \dfrac{y}{\left( \dfrac bcy-b \right)} \\ &\implies \dfrac{a-y}{x} = \dfrac{cy}{by-bc} \\ &\implies aby-abc-by^2+bcy = cxy \\ &\implies by^2 = bcy -cxy +aby - abc \\ &\implies y^2 = cy -\dfrac cb xy +ay - ac \\ \end{align}