Vectors and conic sections - Part 3

Let the coordinates of $P(x_p, y_p)$ and $Q(x_q, y_q)$ . Since $F_1$ is on the directrix and $F_2$ is the focus of the parabola, this means that $x_p+1 = |F_2P|=u$ and $x_q+1 = |F_2Q|=v$ as shown in the figure.

Since $\dfrac {|F_1P|}{|F_1Q|} = \dfrac {x_p+1}{x_q+1} = \dfrac {|F_2P|}{|F_2Q|} = \dfrac uv= \lambda \implies u=\lambda v$ . Also $|F_1P| = \sqrt{(1+x_p)^2+y_p^2} = \sqrt{u^2+4x_p} = \sqrt{u^2+4u-4}$ ; similarly, $|F_1Q| = \sqrt{v^2+4v-4}$ . As $|F_1P| = \lambda |F_1Q|$ , we have:

$\begin{aligned} \sqrt{u^2+4u-4} & = \lambda \sqrt{v^2+4v-4} \\ u^2+4u-4 & = \lambda^2(v^2+4v-4) & \small \blue{\text{Since }u=\lambda v} \\ \lambda^2v^2+4\lambda v-4 & = \lambda^2v^2+4\lambda^2v-4\lambda^2 \\ 4\lambda(\lambda-1)v & = 4(\lambda^2-1) \\ \implies v & = \frac {\lambda+1}\lambda \end{aligned}$

Now $|F_2P| \cdot |F_1Q| = uv = \lambda v^2 = \dfrac {(\lambda+1)^2}\lambda$ $\implies \begin{cases} \lambda = 2 & \implies uv = \frac 92 \\ \lambda = 3 & \implies uv = \frac {16}3 \end{cases}$ $\implies \lfloor 10000 (r-l) \rfloor = \boxed{8333}$ .

Let the equation of $\overline {F_1P}$ be $y=m(x+1)$ , the position coordinates of points $P, Q$ be $(h^2,2h), (k^2,2k)$ respectively. Then the given equation implies

$\lambda^2\left ((k^2+1)^2+4k^2\right )=(h^2+1)^2+4h^2$

Solving the equations of the parabola and the straight line we get

$m^2(x+1)^2-4x=0$ , whose roots are $h^2,k^2$ . So $hk=1\implies k=\dfrac{1}{h}$ .

Therefore $\lambda^2\left ((\frac{1}{h^2}+1)^2+\frac{4}{h^2}\right ) =(h^2+1)^2+4h^2\implies h^2=\lambda$ .

Now, $\vec {F_2P}.\vec {F_2Q}=6-(h^2+\frac{1}{h^2})$

So, $r=6-2.5=3.5, l=6-\frac{10}{3}=\frac{8}{3}=2.666666...$ , and $r-l\approx 0.8333333\implies \lfloor 10000(r-l)\rfloor =\boxed {8333}$