Vectors and conic sections - Part 4

Draw diagonals $F_2A$ and $F_1B$ , and extend $AB$ and $F_1F_2$ to meet at $C$ . Let $p = F_2B$ and $q = AB$ . Since $\overrightarrow{F_1A} = 5\overrightarrow{F_2B}$ , $F_1A = 5p$ and $F_1A || F_2B$ .

By the properties of an ellipse with an equation $\frac{x^2}{3} + y^2 = 1$ , the distance between the foci is $F_1F_2 = 2\sqrt{3 - 1} = 2\sqrt{2}$ , and $F_1A + F_2A = F_1B + F_2B = 2\sqrt{3}$ . Therefore, $F_2A = 2\sqrt{3} - 5d$ and $F_1B = 2\sqrt{3} - d$ .

Since $F_1A || F_2B$ , $\angle F_1AF_2 = \angle AF_2B$ and $\angle F_1BF_2 = \angle AF_1B$ as alternate interior angles.

By the law of cosines on $\triangle F_1AF_2$ and $\triangle AF_2B$ , $\cos \angle F_1AF_2 = \cos \angle AF_2B = \frac{(5p)^2 + (2\sqrt{3} - 5p)^2 - (2\sqrt{2})^2}{2(5p)(2\sqrt{3} - 5p)} = \frac{p^2 + (2\sqrt{3} - 5p)^2 - q^2}{2p(2\sqrt{3} - 5p)}$ , which rearranges to:

$q^2 = 16p^2 - 16\sqrt{3}p + \frac{56}{5}$ .

Similarly, by the law of cosines on $\triangle F_1BF_2$ and $\triangle AF_1B$ , $\cos \angle F_1BF_2 = \cos \angle AF_1B = \frac{p^2 + (2\sqrt{3} - p)^2 - (2\sqrt{2})^2}{2p(2\sqrt{3} - p)} = \frac{(5p)^2 + (2\sqrt{3} - p)^2 - q^2}{2(5p)(2\sqrt{3} - p)}$ , which rearranges to:

$q^2 = 16p^2 + 16\sqrt{3}p - 8$ .

These two equations solve to $p = \frac{\sqrt{3}}{5}$ and $q = \frac{2\sqrt{22}}{5}$ , which means $F_1A = 5p = \sqrt{3}$ and $F_2A = 2\sqrt{3} - 5p = \sqrt{3}$ .

By the law of cosines on $\triangle F_1AF_2$ , $\cos \angle AF_1F_2 = \frac{(2\sqrt{2})^2 + (\sqrt{3})^2 - (\sqrt{3})^2}{2(2\sqrt{2})(\sqrt{3})} = \frac{\sqrt{6}}{3}$ , which means $\sin \angle AF_1F_2 = \sqrt{1 - (\frac{\sqrt{6}}{3})^2} = \frac{\sqrt{3}}{3}$ .

Since $F_1A || F_2B$ , $\triangle ACF_1 \sim \triangle BCF_2$ , so $\frac{F_2C}{F_2B} = \frac{F_1C}{F_1A}$ , or $\frac{F_2C}{p} = \frac{F_2C + 2\sqrt{2}}{5p}$ , which solves to $F_2C = \frac{\sqrt{2}}{2}$ , so $F_1C = F_1F_2 + F_2C = 2\sqrt{2} + \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}$ .

The area of $\triangle ACF_1$ is then $A_{\triangle ACF_1} = \frac{1}{2} \cdot F_1A \cdot F_1C \cdot \sin \angle AF_1F_2 = \frac{1}{2} \cdot \sqrt{3} \cdot \frac{5\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{3} = \frac{5\sqrt{2}}{4}$ .

Since $\triangle ACF_1 \sim \triangle BCF_2$ and ratio of their lengths is $\frac{F_1A}{F_2B} = \frac{5p}{p} = 5$ , the ratio of their areas is $5^2 = 25$ , so the area of $\triangle BCF_2$ is $A_{\triangle BCF_2} = \frac{1}{25}A_{\triangle ACF_1}$ .

That means the area of the quadrilateral $A_{ABF_2F_1} = A_{\triangle ACF_1} - A_{\triangle BCF_2} = A_{\triangle ACF_1} - \frac{1}{25}A_{\triangle ACF_1} = \frac{24}{25}A_{\triangle ACF_1} = \frac{24}{25} \cdot \frac{5\sqrt{2}}{4} = \frac{6\sqrt{2}}{5}$ .

Therefore, $\lfloor 10000 A \rfloor = \lfloor 10000 \cdot \frac{6\sqrt{2}}{5} \rfloor = \boxed{16970}$ .

The given condition $\vec {F_1A}=5\vec {F_2B}$ implies that the lines of action of those two vectors are always parallel. Let the equation of $\overline {F_2B}$ be $y=m(x-\sqrt 2)$ and that of $\overline {F_1A}$ be $y=m(x+\sqrt 2)$ (the position coordinates of $F_1$ are $(-\sqrt 2, 0)$ and of $F_2$ are $(\sqrt 2, 0)$ ).

Then the area of the trapezium $ABF_2F_1$ is $\dfrac{1}{2}(|\overline {AF_1}|+|\overline {BF_2}|)p$ , where $p=\dfrac{2\sqrt 2m}{\sqrt {1+m^2}}$ is the perpendicular distance of $\overline {BF_2}$ from $F_1$ .

That is, the area is $\dfrac{6\sqrt 2m}{\sqrt {1+m^2}}|\overline {BF_2}|$ .

Now, solving the equations of the ellipse and $\overline {AF_1}$ for positive $x$ -value we get the $x$ coordinate of $A$ as $x_A=\dfrac{\sqrt {3(1+m^2)}-3\sqrt 2m^2}{1+3m^2}$

Similarly, we get the $x$ coordinate of $B$ as $x_B=\dfrac{\sqrt {3(1+m^2)}+3\sqrt 2m^2}{1+3m^2}$ .

So, $|\overline {AF_1}|^2=\dfrac{1+m^2}{(1+3m^2)^2}(\sqrt {3(1+m^2)}+\sqrt 2) ^2$ ,

$|\overline {BF_2}|^2=\dfrac{1+m^2}{(1+3m^2)^2}(\sqrt {3(1+m^2)}-\sqrt 2) ^2$ . The given condition $|\overline {AF_1}|=5|\overline {BF_2}|$ yields $m^2=\dfrac{1}{2}\implies |\overline {BF_2}|=\dfrac{\sqrt 3}{5}$ .

Therefore the required area is $\dfrac{6\sqrt 2\times \dfrac{1}{\sqrt 2}}{\sqrt {\dfrac{3}{2}}}\times \dfrac{\sqrt 3}{5}=1.2\times \sqrt 2 \approx 1.697056$ .

Hence the required answer is $\boxed {16970}$ .

Answer to the problem "Vectors and conic sections - Part 2"

Let the position coordinates of $Q$ and $R$ be $(k, m-2k)$ and $(h, m-2h)$ respectively, and the required ratio be $α$ . Then $h=αk$ .

Solving the equations of the hyperbola and the straight line we get $x^2-4mx+m^2+3=0$ , whose roots are $h, k$ .

Then the given equation yields $2m+\sqrt {3(m^2-1)}=2αm-α\sqrt {3(m^2-1)}\implies α^2-14α+1\leq 0\implies α\leq 7+4\sqrt 3$ .

Since the minimum value of $α$ is $1$ when the line touches the hyperbola, the range of $α$ is $[1,7+4\sqrt 3]$ .

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