Vectors and conic sections - Part 2

Points $Q(x_q,y_q)$ and $R(x_r,y_r)$ satisfy both $y = -2x+m$ and $x^2 - \frac {y^2}2 = 1$ . Then we have:

$\begin{aligned} 3 x^2 - y^2 - 3 & = 0 \\ 3x^2 -(4x^2-4mx+m^2) - 3 & = 0 \\ x^2 - 4mx + m^2 + 3 & = 0 \end{aligned}$

$\begin{aligned} \implies x & = \frac {4m\pm \sqrt{16m^2-4m^2-12}}2 = 2m \pm \sqrt{3(m^2-1)} \end{aligned}$

$\implies x_r = 2m + \sqrt{3(m^2-1)}$ and $x_q = 2m - \sqrt{3(m^2-1)}$ . Now we note that $\dfrac {|PR|}{|PQ|} = \dfrac {x_r}{x_q} = \dfrac {2m + \sqrt{3(m^2-1)}}{2m - \sqrt{3(m^2-1)}}$ . Its lowest bound is when $m=1$ , then $\dfrac {|PR|}{|PQ|} = 1$ . Its highest bound is when $m \to \infty$ , then

$\begin{aligned} \frac {|PR|}{|PQ|} & = \lim_{m \to \infty} \frac {2m + \sqrt{3(m^2-1)}}{2m - \sqrt{3(m^2-1)}} = \lim_{m \to \infty} \frac {2 + \sqrt{3\left(1-\frac 1{m^2}\right)}}{2 - \sqrt{3\left(1-\frac 1{m^2}\right)}} = \frac {2+\sqrt 3}{2-\sqrt 3} = 7+ 4 \sqrt 3 \end{aligned}$

Therefore, $\lfloor 1000(r-l) \rfloor = \lfloor 1000(7+4\sqrt 3-1) \rfloor = \boxed{12928}$ .

Substituting $y = -2x + m$ into $x^2 - \frac{y^2}{3} = 1$ gives $x^2 - \frac{(-2x + m)^2}{3} = 1$ , which solves to $x = 2m \pm \sqrt{3m^2 - 3}$ .

This means the $x$ -coordinate of $Q$ is $Q_x = 2m - \sqrt{3m^2 - 3}$ , and using $y = -2x + m$ , the $y$ -coordinate of $Q$ is $Q_y = -2Q_x + m$ , which simplifies to $Q_y = -3m + 2\sqrt{3m^2 - 3}$ .

Similarly, the $x$ -coordinate of $R$ is $R_x = 2m + \sqrt{3m^2 - 3}$ , and using $y = -2x + m$ , the $y$ -coordinate of $R$ is $R_y = -2R_x + m$ , which simplifies to $R_y = -3m - 2\sqrt{3m^2 - 3}$ .

Since $P$ is on the $y$ -axis and on the line $y = -2x + m$ , the coordinates of $P$ are $P_x = 0$ and $P_y = m$ .

The ratio $k = \frac{|PR|}{|PQ|} = \frac{\sqrt{(R_x - P_x)^2 + (R_y - P_y)^2}}{\sqrt{(Q_x - P_x)^2 + (Q_y - P_y)^2}}$ , which simplifies to:

$k = \frac{7m^2 + 4m\sqrt{3m^2 - 3} - 3}{m^2 + 3}$

Since $k$ is an increasing function for $m > 0$ , the lowest value of $k$ occurs when the square root $3m^2 - 3 = 0$ , which is at $m = 1$ and $k = \frac{7 \cdot 1^2 + 4 \cdot 1 \sqrt{3 \cdot 1^2 - 3} - 3}{1^2 + 3} = 1 = l$ , and the highest value of $k$ occurs at $\displaystyle k = \lim_{m \to \infty} \frac{7m^2 + 4m\sqrt{3m^2 - 3} - 3}{m^2 + 3} = \lim_{m \to \infty} \frac{\frac{7m^2}{m^2} + \frac{4m}{m}\sqrt{\frac{3m^2}{m^2} - \frac{3}{m^2}} - \frac{3}{m^2}}{\frac{m^2}{m^2} + \frac{3}{m^2}} = 7 + 4\sqrt{3} = r$ .

Therefore, $r - l = (7 + 4\sqrt{3}) - 1 = 6 + 4\sqrt{3}$ , so $\lfloor 1000(r - l) \rfloor = \lfloor 1000(6 + 4 \sqrt{3}) \rfloor = \boxed{12928}$ .