A tangent is drawn to the ellipse frac{x^2}{2 | vedclass

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(B) Given the ellipse $\frac{x^2}{27} + y^2 = 1$,we have $a^2 = 27$ and $b^2 = 1$,so $a = 3\sqrt{3}$ and $b = 1$.The equation of the tangent at $(a \c

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$\frac{\pi}{6}$

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(B) Given the ellipse $\frac{x^2}{27} + y^2 = 1$,we have $a^2 = 27$ and $b^2 = 1$,so $a = 3\sqrt{3}$ and $b = 1$.The equation of the tangent at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1$.Substituting the values,the equation of the tangent is $\frac{x}{3\sqrt{3}} \cos 	heta + y \sin 	heta = 1$.The $x$-intercept is $3\sqrt{3} \sec 	heta$ and the $y$-intercept is $\csc 	heta$.Let $S$ be the sum of the intercepts: $S = 3\sqrt{3} \sec 	heta + \csc 	heta$.To find the minimum,differentiate $S$ with respect to $	heta$: $\frac{dS}{d	heta} = 3\sqrt{3} \sec 	heta 	an 	heta - \csc 	heta \cot 	heta$.Setting $\frac{dS}{d	heta} = 0$,we get $3\sqrt{3} \frac{\sin 	heta}{\cos^2 	heta} = \frac{\cos 	heta}{\sin^2 	heta}$.This simplifies to $	an^3 	heta = \frac{1}{3\sqrt{3}} = (\frac{1}{\sqrt{3}})^3$.Thus,$	an 	heta = \frac{1}{\sqrt{3}}$,which implies $	heta = \frac{\pi}{6}$.Since the second derivative is positive at $	heta = \frac{\pi}{6}$,the sum of intercepts is minimum at $	heta = \frac{\pi}{6}$.

$A$ tangent is drawn to the ellipse $\frac{x^2}{27} + y^2 = 1$ at the point $(3\sqrt{3} \cos \theta, \sin \theta)$ (where $\theta \in (0, \frac{\pi}{2})$). Then,the value of $\theta$ such that the sum of the intercepts on the axes made by this tangent is minimum,is

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