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(B) The equation of the chord joining points $A(	heta_1)$ and $B(	heta_2)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by:$\frac{x

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$\sqrt{a^2+b^2}$

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(B) The equation of the chord joining points $A(	heta_1)$ and $B(	heta_2)$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is given by:$\frac{x}{a} \cos \left(\frac{	heta_1-	heta_2}{2}ight)-\frac{y}{b} \sin \left(\frac{	heta_1+	heta_2}{2}ight)=\cos \left(\frac{	heta_1+	heta_2}{2}ight)$Since the chord passes through the focus $S(ae, 0)$,we substitute $x=ae$ and $y=0$ into the equation:$\frac{ae}{a} \cos \left(\frac{	heta_1-	heta_2}{2}ight) = \cos \left(\frac{	heta_1+	heta_2}{2}ight)$$e \cos \left(\frac{	heta_1-	heta_2}{2}ight) = \cos \left(\frac{	heta_1+	heta_2}{2}ight)$Using the relation $e = \frac{\sqrt{a^2+b^2}}{a}$,we get:$\frac{\sqrt{a^2+b^2}}{a} \cos \left(\frac{	heta_1-	heta_2}{2}ight) = \cos \left(\frac{	heta_1+	heta_2}{2}ight)$$\sqrt{a^2+b^2} \cos \left(\frac{	heta_1-	heta_2}{2}ight) = a \cos \left(\frac{	heta_1+	heta_2}{2}ight)$Comparing this with the given equation $a \cos \left(\frac{	heta_1+	heta_2}{2}ight) = k \cos \left(\frac{	heta_1-	heta_2}{2}ight)$,we find:$k = \sqrt{a^2+b^2}$

Let $A(\theta_1)$ and $B(\theta_2)$ be two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $S$ be the focus of the hyperbola. If $A, S, B$ are collinear and $a \cos \left(\frac{\theta_1+\theta_2}{2}\right)=k \cos \left(\frac{\theta_1-\theta_2}{2}\right)$ then $k=$

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