SAT1000 - P811

Position coordinates of $M$ and $N$ are $(\dfrac{\sqrt t(3-tk^2)}{3+tk^2},\dfrac{6k\sqrt t}{3+tk^2})$ and

$(\dfrac{\sqrt t(3k^2-t)}{3k^2+t},-\dfrac{6k\sqrt t}{3k^2+t})$ .

The given condition yields $\dfrac{k}{3k^2+t}=\dfrac{2}{tk^2+3}$

$\implies t=\dfrac{3k(2k-1)}{k^3-2}>3$

$\implies \sqrt[3] 2<k<2$ .

So, $l=\sqrt[3] 2\approx 1.25992, r=2$ and the required answer is $\boxed {2740}$ .

Solution to this problem :

Let the equation of $\overline {AB}$ be $y=mx+c$

Then $3x^2+4(mx+c)^2=12\implies (4m^3)x^2+8mcx+4c^2-12=0$

If the position coordinates of $A$ and $B$ be $(x_1,y_1)$ and $(x_2,y_2)$ respectively, then

$\dfrac{x_1+x_2}{2}=-\dfrac {4mc}{4m^2+3},\dfrac {y_1+y_2}{2}=\dfrac {3c}{4m^2+3}$ .

Since this point is on the line $x=2y, 6c=-4mc\implies m=-\dfrac {3}{2}=-1.5$ .

Area of $\triangle {ABP}$ is $\dfrac{1}{6}(4-c)\sqrt {36-3c^2}$

This attains a maximum when $c\approx -1.64575\implies m+c\approx -3.14575$

Hence $\lfloor 1000(m+c)\rfloor =\boxed {-3146}$ .