From any point on the line `(t+2)(x+y) =1, t ne -2`, tangents are drawn to the ellipse `4x^(2)+16y^(2) = 1`. It is given that chord of contact passes through a fixed point. Then the number of integral values of 't' for which the fixed point always lies inside the ellipse is

Options

A. 0
B. 1
C. 2
D. 3

Answer

C

Explanation

Any point on `(t+2) (x+y) =1` is `(alpha, (1)/(t+2) -alpha)`
Equation of chord of contact is `4xalpha + 16y ((1)/(t+2)-alpha) =1`
`rArr alpha (4x-16y) + ((16)/(t+2)y-1) =0`
The chord of contact passes through the intersection of
`x - 4y = 0` and `((16)/(t+2)y -1) =0`
`:.` The point of intersection is `((t+2)/(4),(t+2)/(16))`
It lies inside the ellipse if `4 ((t+2)/(4))^(2)+16 ((t+2)/(16))^(2) -1 lt 0`
`rArr (t+2)^(2) lt (16)/(5)`
`rArr (-(4)/(sqrt(5))-2)lt t lt ((4)/(sqrt(5))-2)`
`rArr t =- 3, -1`. (as `t ne -2)`