If the angle between the asymptotes of the hy | vedclass

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(D) The equation of the hyperbola is $x^2-k y^2=3$,which can be written as $\frac{x^2}{3}-\frac{y^2}{3/k}=1$. Here $a^2=3$ and $b^2=\frac{3}{k}$. The

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$\left(k,-\frac{\sqrt{3} e}{2}\right)$

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(D) The equation of the hyperbola is $x^2-k y^2=3$,which can be written as $\frac{x^2}{3}-\frac{y^2}{3/k}=1$. Here $a^2=3$ and $b^2=\frac{3}{k}$. The angle between the asymptotes is $2	an^{-1}\left(\frac{b}{a}ight) = \frac{\pi}{3}$,so $	an^{-1}\left(\frac{b}{a}ight) = \frac{\pi}{6}$. Thus,$\frac{b}{a} = 	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$. Squaring gives $\frac{b^2}{a^2} = \frac{1}{3}$ $\Rightarrow \frac{3/k}{3} = \frac{1}{3}$ $\Rightarrow \frac{1}{k} = \frac{1}{3}$ $\Rightarrow k=3$. The hyperbola is $\frac{x^2}{3}-y^2=1$. Eccentricity $e = \sqrt{1+\frac{b^2}{a^2}} = \sqrt{1+\frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$. The pole of the line $lx+my+n=0$ with respect to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\left(-\frac{a^2l}{n}, \frac{b^2m}{n}ight)$. For $x+y-1=0$,$l=1, m=1, n=-1$. Pole is $\left(-\frac{3(1)}{-1}, \frac{1(1)}{-1}ight) = (3, -1)$. Since $k=3$ and $e=\frac{2}{\sqrt{3}}$,we have $-1 = -\frac{\sqrt{3}}{2} 	imes \frac{2}{\sqrt{3}} = -\frac{\sqrt{3}e}{2}$. Thus,the pole is $\left(k, -\frac{\sqrt{3}e}{2}ight)$.

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

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