Determine possible equations of line $r$, given $PA \cdot PB=17$

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Consider a line $r$ that passes through the point $P(2, 3)$. The line $r$ intersects the curve $x^2-2xy-y^2=0$ at points $A$ and $B$. Knowing that the geometric locus defined by the curve is determined by two perpendicular lines, determine the possible equations of line $r$, given that $PA \cdot PB=17$.

Solution: Is there any other way to solve the problem more easily, using knowledge of analytic geometry with a vectorial approach? Or calculus...? Below is the procedure that would be very effective...

$$ \begin{aligned} & \text { Coordinates of } A \text { : } \\ & \left\{\begin{array}{l} y=(\sqrt{2}-1) x \\ y=m x-2 m+3 \end{array}\right. \end{aligned} $$

$$ x=\frac{3-2 m}{\sqrt{2}-1-m} \text { e } y=\frac{3 \sqrt{2}-3-2 m \sqrt{2}+2 m}{\sqrt{2}-1-m} $$ Coordinates of $B$ : $$ \begin{aligned} & \left\{\begin{array}{l} y=-(\sqrt{2}+1) x \\ y=m x-2 m+3 \end{array}\right. \\ & x=\frac{2 m-3}{\sqrt{2}+1+m} \text { e } y=\frac{3 \sqrt{2}+3-2 m \sqrt{2}-2 m}{\sqrt{2}+1+m} \end{aligned} $$

$$ \begin{aligned} & \sqrt{\left(\frac{3-2 m}{\sqrt{2}-1-m}-2\right)^2+\left(\frac{3 \sqrt{2}-3-2 m \sqrt{2}+2 m}{\sqrt{2}-1-m}-3\right)^2} \sqrt{\left(\frac{2 m-3}{\sqrt{2}+1+m}-2\right)^2+\left(\frac{3 \sqrt{2}+3-2 m \sqrt{2}-2 m}{\sqrt{2}+1+m}-3\right)^2}=17 \\ \end{aligned} $$

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Problems involving sum or products of two or more lengths are usually tackled in analytic geometry via parametric coordinates.

Recall a line inclined at $\theta$ to positive $x$ axis and passing through $P=(2,3)$ has parametric form $(2+r\cos \theta, 3+r\sin \theta)$ where $r$ is signed distance. Putting these coordinates in the curve equation $x^2-2xy-y^2=0$, gives $$r^2(\cos 2\theta-\sin 2\theta)-2r(\cos\theta + 5\sin \theta)-17=0$$ which is a quadratic in $r$ with roots $|PA|, |PB|$ (signed). Product of roots is $$|PA|\cdot |PB|=\frac{-17}{\cos 2\theta - \sin 2\theta}=\pm \,17 \quad \text{(given)}$$ $$\cos 2\theta - \sin 2\theta = \pm 1 \; \Rightarrow \; \theta = 0^\circ, 45^\circ, 90^\circ, 135^\circ $$ Hence four lines are obtained in total $$y=3 \quad \text{or} \quad y=x+1 \quad \text{or} \quad x=2 \quad \text{or}\quad x+y=5$$

Verification for two of these lines to satisfy $PA\cdot PB=17$ in Geogebra :

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