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

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(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a}

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$\pi/6$

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(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.Here $a^2 = 27 \implies a = 3\sqrt{3}$ and $b^2 = 1 \implies b = 1$.So,the tangent equation is $\frac{x \cos 	heta}{3\sqrt{3}} + y \sin 	heta = 1$.The $x$-intercept is $a_x = \frac{3\sqrt{3}}{\cos 	heta} = 3\sqrt{3} \sec 	heta$ and the $y$-intercept is $a_y = \frac{1}{\sin 	heta} = \csc 	heta$.The sum of intercepts is $f(	heta) = 3\sqrt{3} \sec 	heta + \csc 	heta$.To find the minimum,set $f'(	heta) = 3\sqrt{3} \sec 	heta 	an 	heta - \csc 	heta \cot 	heta = 0$.$3\sqrt{3} \frac{\sin 	heta}{\cos^2 	heta} = \frac{\cos 	heta}{\sin^2 	heta} \implies 	an^3 	heta = \frac{1}{3\sqrt{3}} = \left(\frac{1}{\sqrt{3}}ight)^3$.Thus,$	an 	heta = \frac{1}{\sqrt{3}}$,which gives $	heta = \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 $\theta \in (0, \pi/2)$. The value of $\theta$ for which the sum of the intercepts on the axes made by this tangent is minimum,is:

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