A tangent is drawn to the ellipse `(x^2)/(27)+y^2=1` at `(3sqrt(3)costheta(0,pi/2)dot` Then find the value of `theta` such that the sum of intercepts on the axes made by this tangent is minimum.

Options

A. `(pi)/(3)`
B. `(pi)/(6)`
C. `(pi)/(8)`
D. `(pi)/(4)`

Answer

2

Explanation

Equation of the tangenth to the ellipese `(x^(2))/(27)+y^(2) =1 at 3sqrt(3 cos theta sin theta),theta in (0,pi//2)` is
`(sqrt(3)cos theta)/(9)+y sin theta=1`
`therefore` sum of the intercepts =S =`3sqrt(3)theta+cosec theta`
For minimum values of s,`(ds)/(d theta)=0`
`3(sqrt(3)sin theta)/(cos^(2)theta)-(cos theta)/(sin^(2)theta)=0`
or `3sqrt(3)sin^(3) theta - cos^(3) theta =0`
or `tan theta =(1)sqrt(3)=tan pi//6 rarr theta -pi//6`