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(B) The equations are $\frac{x^{2}}{9}+y^{2}=1$ and $x^{2}+y^{2}=3$. Subtracting the equations: $x^{2}(1 - \frac{1}{9}) = 3 - 1 \implies \frac{8}{9}x^

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$\frac{2}{\sqrt{3}}$

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(B) The equations are $\frac{x^{2}}{9}+y^{2}=1$ and $x^{2}+y^{2}=3$. Subtracting the equations: $x^{2}(1 - \frac{1}{9}) = 3 - 1 \implies \frac{8}{9}x^{2} = 2 \implies x^{2} = \frac{9}{4} \implies x = \frac{3}{2}$ (in the first quadrant).Substituting $x^{2} = \frac{9}{4}$ into $x^{2}+y^{2}=3$,we get $y^{2} = 3 - \frac{9}{4} = \frac{3}{4} \implies y = \frac{\sqrt{3}}{2}$.The point of intersection is $P(\frac{3}{2}, \frac{\sqrt{3}}{2})$.For the ellipse $\frac{x^{2}}{9}+y^{2}=1$,differentiating gives $\frac{2x}{9} + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{9y}$.At $P$,$m_{1} = -\frac{3/2}{9(\sqrt{3}/2)} = -\frac{3}{9\sqrt{3}} = -\frac{1}{3\sqrt{3}}$.For the circle $x^{2}+y^{2}=3$,differentiating gives $2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$.At $P$,$m_{2} = -\frac{3/2}{\sqrt{3}/2} = -\sqrt{3}$.The angle $	heta$ between the tangents is given by $	an 	heta = |\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}|$.$	an 	heta = |\frac{-1/(3\sqrt{3}) - (-\sqrt{3})}{1 + (-1/(3\sqrt{3}))(-\sqrt{3})}| = |\frac{-1/(3\sqrt{3}) + \sqrt{3}}{1 + 1/3}| = |\frac{(-1+9)/(3\sqrt{3})}{4/3}| = |\frac{8}{3\sqrt{3}} 	imes \frac{3}{4}| = \frac{2}{\sqrt{3}}$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to:

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