Question

Tangent is drawn to the ellipse `x^2/27 + y^2 = 1` at the point `(3sqrt(3) cos theta, sin theta), 0 < 0 < 1`. The sum of the intercepts on the axes made by the tangent is minimum if 0 is equal to

Options

• `pi/6`

• `pi/4`

• `pi/3`

• `(3pi)/4`

Answer 1

`pi/6`

Explanation:

∵ `x^2/27 + y^2` = 1

⇒ `(2x)/27 + 2y (dy)/(dx)` = 0

⇒ `(dy)/(dx) = (-x)/(27y) = (-3sqrt(3) cos theta)/(27 sin theta)`

Equation of tangent

`y - sin theta = (-3sqrt(3) cos theta)/(27 sin theta) (x - 3sqrt(3)  cos theta)`

`27 sin theta y - 27 sin^2theta = 3sqrt(3) cos thetax + 27 cos^2theta`

`27 sin theta y + 3sqrt(3) cos thetax` = 27

x intercept = `(3sqrt(3))/cos theta`, y intercept = `1/sintheta`

sum, s = `(3sqrt(3))/cos theta + 1/sintheta = 3sqrt(3) sec theta + "cosec"  theta`

`(ds)/(d theta) = 3sqrt(3) sec theta tan theta - "cosec" theta . cot theta`

= `(3sqrt(3) sin theta)/(cos^2 theta) - cos theta/(sin^2 theta)` = 0

⇒ `tan^2 theta = 1/(3sqrt(3)`

⇒ `tan theta = 1/sqrt(3)`

`theta = pi/6, (7pi)/6` ...

`(d^2s)/(d theta^2) = 3sqrt(3)  (sec theta. tan^2 theta ++ sec^3 theta + "cosec" theta. cot^2 theta + "cosec"^3 theta)`

∴ `((d^2s)/(d theta^2)) (theta = pi/6) > 0`

∴ Sum is minimum at `theta = pi/6`