SAT1000 - P759

Equations of asymptotes of given hyperbola is $\displaystyle y = \frac{b}{a}x$ and $\displaystyle y = -\frac{b}{a}x$ .

Points $A$ and $B$ can be determined by the intersection of each asymptote with the line: $\displaystyle x - 3y + m = 0$ . We get

$\displaystyle A = \bigg(\frac{-ma}{3b + a},\frac{mb}{3b + a}\bigg)\hspace{50pt} B = \bigg(\frac{ma}{3b - a},\frac{mb}{3b + a}\bigg)$

Point $P$ is at $\displaystyle (m,0)$ .

$\displaystyle PA^2 = \bigg(m - \frac{-ma}{3b + a}\bigg)^2 + \bigg(0 - \frac{mb}{3b + a}\bigg)^2$

$\displaystyle \Rightarrow PA^2 = m^2 + \frac{m^2a^2}{(3b + a)^2} + \frac{2m^2a}{3b + a} + \frac{m^2b^2}{(3b + a)^2}$

$\displaystyle \Rightarrow PA^2 = m^2\bigg(\frac{(3b + a)^2 + a^2 + 2a(3b + a) + b^2}{(3b + a)^2}\bigg)$

$\displaystyle\Rightarrow PA^2 = \frac{m^2(4a^2 + 10b^2 + 12ab)}{(3b + a)^2}\hspace{35pt}\cdots\hspace{15pt}\text{Eq. 1}$

Similarly, writing for $PB^2$

$\displaystyle PB^2 = \bigg(m - \frac{ma}{3b - a}\bigg)^2 + \bigg(0 - \frac{mb}{3b - a}\bigg)^2$

$\displaystyle \Rightarrow PB^2 = m^2 + \frac{m^2a^2}{(3b - a)^2} - \frac{2m^2a}{3b - a} + \frac{m^2b^2}{(3b - a)^2}$

$\displaystyle \Rightarrow PB^2 = m^2\bigg(\frac{(3b - a)^2 + a^2 - 2a(3b - a) + b^2}{(3b - a)^2}\bigg)$

$\displaystyle\Rightarrow PB^2 = \frac{m^2(4a^2 + 10b^2 - 12ab)}{(3b - a)^2}\hspace{35pt}\cdots\hspace{15pt}\text{Eq. 2}$

It is given that $|PA| = |PB|$

$\displaystyle\Rightarrow PA^2 = PB^2$

Putting the expressions of $PA^2$ and $PB^2$ from Eq. 1 and Eq. 2 in the above equation we get

$\displaystyle \frac{m^2(4a^2 + 10b^2 + 12ab)}{(3b + a)^2} = \frac{m^2(4a^2 + 10b^2 - 12ab)}{(3b - a)^2}$

$\displaystyle \Rightarrow \frac{2a^2 + 5b^2 + 6ab}{(3b + a)^2} = \frac{2a^2 + 5b^2 - 6ab}{(3b - a)^2}$

$\displaystyle \Rightarrow \frac{(3b + a)^2}{2a^2 + 5b^2 + 6ab} = \frac{(3b - a)^2}{2a^2 + 5b^2 - 6ab}\hspace{25pt}[$ Reciprocating both sides $]$

$\displaystyle \Rightarrow \frac{a^2 + 9b^2 + 6ab}{2a^2 + 5b^2 + 6ab} = \frac{a^2 + 9b^2 - 6ab}{2a^2 + 5b^2 - 6ab}$

$\displaystyle \Rightarrow 1 + \frac{4b^2 - a^2}{2a^2 + 5b^2 + 6ab} = 1 + \frac{4b^2 - a^2}{2a^2 + 5b^2 - 6ab}$

$\displaystyle \Rightarrow \frac{4b^2 - a^2}{2a^2 + 5b^2 + 6ab} = \frac{4b^2 - a^2}{2a^2 + 5b^2 - 6ab}$

Above equation is true only if numerator is $0$ or denominators are equal. Denominator can be equal when any one of $a$ or $b$ is $0$ which is not possible for existence of hyperbola. So numerator must be $0$ .

$\displaystyle \Rightarrow 4b^2 - a^2 = 0$

$\displaystyle \Rightarrow \frac{a^2}{b^2} = 4 \Rightarrow \frac{b^2}{a^2} = \frac{1}{4}$

Eccentricity of hyperbola is given by $\displaystyle E = \sqrt{1 + \frac{b^2}{a^2}}$

So, $\displaystyle E = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$

$\displaystyle \Rightarrow 1000E = 1000\frac{\sqrt{5}}{2} = 500\sqrt{5} \approx 1118.034$

$\displaystyle \Rightarrow \lfloor 1000E \rfloor = 1118$