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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Tangent is drawn to the ellipse $\large\frac{x^2}{27} $$+y^2=1$ at $(3 \sqrt 3 \cos \theta, \sin \theta)$ (where $\theta \in (0, \large\frac{\pi}{2} )$) Then the value of $\theta$ such that sum of intercepts on axis made by this tangent is minimum is

$\begin{array}{1 1}(A)\;\frac{\pi}{3} \\(B)\;\frac{\pi}{6} \\(C)\;\frac{\pi}{8} \\(D)\;\frac{\pi}{4} \end{array}$

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Equation of tangent at $(3 \sqrt 3 \cos \theta , \sin \theta)$ is $\large\frac{x (3 \sqrt 3 \cos \theta)}{27} +\frac{y(\sin \theta)}{1}$$=1$
$\large\frac{x \cos \theta}{3 \sqrt 3}$$+y \sin \theta=1$
$\therefore$ Sum of intersects on axes.
$\qquad= 3 \sqrt 3 \sec \theta + cosec \theta=f(\theta)$
$\therefore f'(\theta)=\large\frac{3 \sqrt 3 \sin ^3 \theta- \cos ^3 \theta}{\sin ^2 \theta \cos ^2 \theta}$
For $f(\theta)$ to be minimum $f'(\theta)=0$
=> $ 3 \sqrt 3 \sin ^3 \theta = \cos ^3 \theta$
=> $ \tan \theta =\large\frac{1}{\sqrt 3}$
$\theta =\large\frac{\pi}{6}$
Hence B is the correct answer.
answered Apr 10, 2014 by meena.p

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