For some theta in (0, frac{pi}{2}), let the e | vedclass

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(C) For the hyperbola $x^{2} - y^{2} \sec^{2} 	heta = 8$, we rewrite it as $\frac{x^{2}}{8} - \frac{y^{2}}{8 \cos^{2} 	heta} = 1$. Here $a^{2} = 8$

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$8$

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(C) For the hyperbola $x^{2} - y^{2} \sec^{2} 	heta = 8$, we rewrite it as $\frac{x^{2}}{8} - \frac{y^{2}}{8 \cos^{2} 	heta} = 1$. Here $a^{2} = 8$ and $b^{2} = 8 \cos^{2} 	heta$. $e_{1} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \cos^{2} 	heta}$. $l_{1} = \frac{2b^{2}}{a} = \frac{2(8 \cos^{2} 	heta)}{\sqrt{8}} = 4 \sqrt{2} \cos^{2} 	heta$. For the ellipse $x^{2} \sec^{2} 	heta + y^{2} = 6$, we rewrite it as $\frac{x^{2}}{6 \cos^{2} 	heta} + \frac{y^{2}}{6} = 1$. Here $a^{2} = 6$ and $b^{2} = 6 \cos^{2} 	heta$. $e_{2} = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \cos^{2} 	heta} = \sin 	heta$. $l_{2} = \frac{2b^{2}}{a} = \frac{2(6 \cos^{2} 	heta)}{\sqrt{6}} = 2 \sqrt{6} \cos^{2} 	heta$. Given $e_{1}^{2} = e_{2}^{2}(\sec^{2} 	heta + 1)$, we have $1 + \cos^{2} 	heta = \sin^{2} 	heta (1 + \frac{1}{\cos^{2} 	heta}) = \sin^{2} 	heta (\frac{\cos^{2} 	heta + 1}{\cos^{2} 	heta}) = 	an^{2} 	heta (1 + \cos^{2} 	heta)$. Since $1 + \cos^{2} 	heta 
eq 0$, we get $	an^{2} 	heta = 1$, so $	heta = \frac{\pi}{4}$. At $	heta = \frac{\pi}{4}$, $\cos^{2} 	heta = \frac{1}{2}$, $e_{1} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$, $l_{1} = 4 \sqrt{2} (\frac{1}{2}) = 2 \sqrt{2}$. $e_{2} = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, $l_{2} = 2 \sqrt{6} (\frac{1}{2}) = \sqrt{6}$. Then $(\frac{l_{1}l_{2}}{e_{1}e_{2}}) 	an^{2} 	heta = (\frac{2 \sqrt{2} \cdot \sqrt{6}}{\sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}}}) \cdot 1 = \frac{2 \sqrt{12}}{\frac{\sqrt{3}}{2}} = \frac{4 \cdot 2 \sqrt{3}}{\sqrt{3}} = 8$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

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