If beta is an angle between the normals drawn | vedclass

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(C) The given curve 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{

Vedclass · Accepted Answer

$\sqrt{3} \cot \beta = \sin 2 	heta$

Vedclass · Answer

(C) The given curve is $x^2 + 3y^2 = 9$. Differentiating with respect to $x$,we get $2x + 6y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{x}{3y}$.At point $P(3 \cos 	heta, \sqrt{3} \sin 	heta)$,the slope of the tangent is $m_{T1} = -\frac{3 \cos 	heta}{3 \sqrt{3} \sin 	heta} = -\frac{1}{\sqrt{3}} \cot 	heta$. The slope of the normal is $m_{N1} = \sqrt{3} 	an 	heta$.At point $Q(-3 \sin 	heta, \sqrt{3} \cos 	heta)$,the slope of the tangent is $m_{T2} = -\frac{-3 \sin 	heta}{3 \sqrt{3} \cos 	heta} = \frac{1}{\sqrt{3}} 	an 	heta$. The slope of the normal is $m_{N2} = -\sqrt{3} \cot 	heta$.The angle $\beta$ between the normals is given by $	an \beta = |\frac{m_{N1} - m_{N2}}{1 + m_{N1} m_{N2}}|$.$	an \beta = |\frac{\sqrt{3} 	an 	heta - (-\sqrt{3} \cot 	heta)}{1 + (\sqrt{3} 	an 	heta)(-\sqrt{3} \cot 	heta)}| = |\frac{\sqrt{3}(	an 	heta + \cot 	heta)}{1 - 3}| = |\frac{\sqrt{3}(\frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta})}{-2}| = \frac{\sqrt{3}}{2 \sin 	heta \cos 	heta} = \frac{\sqrt{3}}{\sin 2 	heta} = \sqrt{3} \operatorname{cosec} 2 	heta$.Thus,$\cot \beta = \frac{1}{\sqrt{3}} \sin 2 	heta$. However,checking the options,we see $	an \beta = \sqrt{3} \operatorname{cosec} 2 	heta$. Re-evaluating the options,option $B$ is $\cot \beta = \sqrt{3} \operatorname{cosec} 2 	heta$. Wait,$	an \beta = \sqrt{3} \operatorname{cosec} 2 	heta$ implies $\cot \beta = \frac{1}{\sqrt{3}} \sin 2 	heta$. Let's re-check the calculation: $	an \beta = \frac{\sqrt{3}}{\sin 2 	heta}$. Therefore $\cot \beta = \frac{\sin 2 	heta}{\sqrt{3}}$. None of the options match exactly,but based on standard problems of this type,the intended answer is often related to $\cot \beta = \frac{1}{\sqrt{3}} \sin 2 	heta$.

If $\beta$ is an angle between the normals drawn to the curve $x^2+3y^2=9$ at the points $P(3 \cos \theta, \sqrt{3} \sin \theta)$ and $Q(-3 \sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \frac{\pi}{2})$,then:

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