A tangent is drawn at (3 sqrt{3} cos theta, s | vedclass

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(A) The equation of the tangent at $(3 \sqrt{3} \cos 	heta, \sin 	heta)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ is given by $\frac{x(3 \sqr

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$\frac{\pi}{6}$

Vedclass · Answer

(A) The equation of the tangent at $(3 \sqrt{3} \cos 	heta, \sin 	heta)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$ is given by $\frac{x(3 \sqrt{3} \cos 	heta)}{27} + \frac{y \sin 	heta}{1} = 1$,which simplifies to $\frac{x \cos 	heta}{3 \sqrt{3}} + y \sin 	heta = 1$. The $x$-intercept is $a = 3 \sqrt{3} \sec 	heta$ and the $y$-intercept is $b = \operatorname{cosec} 	heta$. The sum of the intercepts is $L(	heta) = 3 \sqrt{3} \sec 	heta + \operatorname{cosec} 	heta$. To find the minimum,we differentiate with respect to $	heta$: $\frac{dL}{d	heta} = 3 \sqrt{3} \sec 	heta 	an 	heta - \operatorname{cosec} 	heta \cot 	heta$. Setting $\frac{dL}{d	heta} = 0$,we get $3 \sqrt{3} \frac{\sin 	heta}{\cos^2 	heta} = \frac{\cos 	heta}{\sin^2 	heta}$,which 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 = \frac{\pi}{6}$.

List-$I$	List-$II$
$(P)$ Directrix corresponding to the focus $(-3, 0)$	$(1)$ $y = 4$
$(Q)$ Tangent at the vertex $(0, 4)$	$(2)$ $3x = 25$
$(R)$ Latus rectum through $(3, 0)$	$(3)$ $x = 3$
	$(4)$ $y + 4 = 0$
	$(5)$ $x + 3 = 0$
	$(6)$ $3x + 25 = 0$

$A$ tangent is drawn at $(3 \sqrt{3} \cos \theta, \sin \theta)$ $\left(0 < \theta < \frac{\pi}{2}\right)$ to the ellipse $\frac{x^2}{27}+\frac{y^2}{1}=1$. The value of $\theta$ for which the sum of the intercepts on the coordinate axes made by this tangent attains the minimum is

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