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

B

Explanation

the equation of the tangent at `( 3 sqrt(3) cos theta , sin , theta )` to the ellipse `(x^(2))/(27) +y^(2)=1` is
`((3sqrt(3)cos theta )/(27))x + ( sin theta ) y=1implies ( x cos theta)/(3sqrt(3))+ y sin theta =1`
Let S be the sum of the intercepts made by this tangent on the coordinate axes , then ,
`s=3sqrt(3) sec theta + cosec theta `
`implies ( ds)/(d theta )=3sqrt(3) sec theta tan theta - cosec theta cot theta `
Now, `(ds)/(d theta) =0`
`implies 3sqrt(3) ( sin theta )/( cos^(2) theta ) -( cos theta )/( sin ^(2) theta )=0`
`implies tan^(3) theta (1)/(3sqrt(3)) implies tan theta =(1)/(sqrt(3)) implies theta =(pi)/(6)`
clearly `,((d^(2)s)/(d theta^(2)))_( theta = pi//6)`
hence ,S in maximum for `theta = pi//6.`