If beta is one of the angles between the norm | vedclass

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(B) The equation of the ellipse is $x^2 + 3y^2 = 9$.Differentiating with respect to $x$,we get $2x + 6y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx

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

Vedclass · Answer

(B) The equation of the ellipse is $x^2 + 3y^2 = 9$.Differentiating with respect to $x$,we get $2x + 6y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{3y}$.The slope of the normal at any point $(x, y)$ is $m = -\frac{dx}{dy} = \frac{3y}{x}$.For the point $P_1 = (3\cos 	heta, \sqrt{3} \sin 	heta)$,the slope of the normal $m_1 = \frac{3(\sqrt{3} \sin 	heta)}{3 \cos 	heta} = \sqrt{3} 	an 	heta$.For the point $P_2 = (-3\sin 	heta, \sqrt{3} \cos 	heta)$,the slope of the normal $m_2 = \frac{3(\sqrt{3} \cos 	heta)}{-3 \sin 	heta} = -\sqrt{3} \cot 	heta$.The angle $\beta$ between the normals is given by $	an \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an \beta = \left| \frac{\sqrt{3} 	an 	heta - (-\sqrt{3} \cot 	heta)}{1 + (\sqrt{3} 	an 	heta)(-\sqrt{3} \cot 	heta)} ight| = \left| \frac{\sqrt{3}(	an 	heta + \cot 	heta)}{1 - 3} ight| = \left| \frac{\sqrt{3}(	an 	heta + \cot 	heta)}{-2} ight| = \frac{\sqrt{3}}{2} (	an 	heta + \cot 	heta)$.Since $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\frac{1}{2} \sin 2	heta} = \frac{2}{\sin 2	heta}$.Thus,$	an \beta = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sin 2	heta} = \frac{\sqrt{3}}{\sin 2	heta}$.Therefore,$\frac{1}{\cot \beta} = \frac{\sqrt{3}}{\sin 2	heta}$,which implies $\frac{\cot \beta}{\sin 2	heta} = \frac{1}{\sqrt{3}}$.Hence,$\frac{2 \cot \beta}{\sin 2	heta} = \frac{2}{\sqrt{3}}$.

$List-I$	$List-II$
$(I)$ If $\phi=\frac{\pi}{4}$,then the area of the triangle $FGH$ is	$(P) \frac{(\sqrt{3}-1)^4}{8}$
$(II)$ If $\phi=\frac{\pi}{3}$,then the area of the triangle $FGH$ is	$(Q) 1$
$(III)$ If $\phi=\frac{\pi}{6}$,then the area of the triangle $FGH$ is	$(R) \frac{3}{4}$
$(IV)$ If $\phi=\frac{\pi}{12}$,then the area of the triangle $FGH$ is	$(S) \frac{1}{2\sqrt{3}}$
	$(T) \frac{3\sqrt{3}}{2}$

If $\beta$ is one of the angles between the normals to the ellipse $x^2 + 3y^2 = 9$ at the points $(3\cos \theta, \sqrt{3} \sin \theta)$ and $(-3\sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \pi/2)$,then $\frac{2 \cot \beta}{\sin 2\theta}$ is equal to

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