Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App