If the angle between the asymptotes of the hyperbola $x^2-k y^2=3$ is $\frac{\pi}{3}$ and $e$ is its…
If the angle between the asymptotes of the hyperbola $x^2-k y^2=3$ is $\frac{\pi}{3}$ and $e$ is its eccentricity, then the pole of the line $x+y-1=0$ with respect to this hyperbola is
$\left(k, \frac{\sqrt{3} e}{2}\right)$
$\left(-k, \frac{\sqrt{3} e}{2}\right)$
$\left(-k,-\frac{\sqrt{3} e}{2}\right)$
$\left(k,-\frac{\sqrt{3} e}{2}\right)$
Solution
Given the equation of hyperbola is $x^2-k y^2=3$
$\Rightarrow \frac{x^2}{3}-\frac{y^2}{\frac{3}{k}}=1$
So, angle between the asymptotes is
$\theta=\tan ^{-1}\left(\frac{2 a b}{a^2-b^2}\right)$
$\Rightarrow \tan \left(\frac{\pi}{3}\right)=\frac{2 \times \sqrt{3} \times \frac{\sqrt{3}}{\sqrt{k}}}{3-\frac{3}{k}}$
$\begin{aligned} & \Rightarrow \sqrt{3}=\frac{2 \sqrt{k}}{k-1} \Rightarrow 3=\frac{4 k}{(k-1)^2} \Rightarrow k=3, \frac{1}{3} \\ & \text { So, } e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{3+1}}{\sqrt{3}}=\frac{2}{\sqrt{3}}\end{aligned}$
Now, the hyperbola is $\frac{x^2}{3}-\frac{y^2}{1}=1$
So, pole of the line $x+y-1=0$ with respect to the hyperbola is
$\left(\frac{-a^2 l}{n}, \frac{b^2 m}{n}\right)$
$\begin{aligned} & =\left(\frac{-3 \times 1}{-1}, \frac{1 \times 1}{-1}\right)=(3,-1) \\ & =\left(3,-\frac{\sqrt{3}}{2} \times \frac{2}{\sqrt{3}}\right)=\left(k,-\frac{\sqrt{3}}{2} e\right)\end{aligned}$