Hyperbola

The equation of the hyperbola is$x^2-k y^2=3$which can be rewritten in standard form as:$\;\frac{x^2}{3}-\;\frac{y^2}{\frac{3}{k}}=1$Here, $a^2=3$ and $b^2=\;\frac{3}{k}$.The angle between the asymptotes is given by$\tan ^{-1} \;\frac{b}{a}=\;\frac{\pi}{3}$This implies:$\tan \;\frac{\pi}{6}=\;\frac{b}{a}=\;\frac{\sqrt{\frac{3}{k}}}{\sqrt{3}}=\;\frac{1}{\sqrt{k}}$Therefore:$\;\frac{1}{\sqrt{3}}=\;\frac{1}{\sqrt{k}} \Rightarrow k=3$Recalculate the standard form with $k=3$ :$3 x^2-y^2=9$The pole of the line $x+y-1=0$ with respect to the hyperbola $3 x^2-y^2=9$ can be found. Let $(h, k)$ be $t$ pole. The equation of the pole is defined using $S_1=0$ :$3 h x-k y=9$This equation is rewritten as:$\;\frac{h x}{3}-\;\frac{k y}{3}=1$Comparing with the equation $x+y-1=0$, we get:$\;\frac{h}{3}=1 \;\; \;\text{ and }\; \;\; \;\frac{-k}{3}=1$Solving these gives:$h=3 \;\; \;\text{ and }\; \;\; k=-9$Thus, the pole of the line is $(3,-9)$. The final expression representing the pole in terms of a parameter is:$\left(k, \;\frac{-\sqrt{3} e}{2}\right)$where $k=3$.