From a point `P(1,2)` , two tangents are drawn to a hyperbola `H` in which one tangent is drawn to each arm of the hyperbola. If the equations of the asymptotes of hyperbola `H` are `sqrt(3)x-y+5=0` and `sqrt(3)x+y-1=0` , then the eccentricity of `H` is 2 (b) `2/(sqrt(3))` (c) `sqrt(2)` (d) `sqrt(3)`

Explanation

We know that if angle between asymptotes is `theta,` then eccentricity of hyparbola is `sec.(theta)/(2)`.
A cute angle between asmytotes `sqrt3x-y+5=0` and `sqrt3x+y-1=0" is "(pi)/(3)`.
Let `L_(1)(x,y)=sqrt3x-y+5 and L_(2)(x,y)=sqrt3x+y-1`.
`therefore" "L_(1)(0,0)=5 and L_(2)(0,0)=-1`
Also, `a_(1)a_(2)+b_(1)b_(2)=(sqrt3)(sqrt3)+(-1)(1)=2gt0`.
Thus, origin lies in acute angle.
Now, `L_(1)(1,2)gt0 and L_(2)(1,2)gt0`
So, (1,2) lies in obtuse angle formed by asymptotes.
Since tangents drawn from point (1, 2) are to different branches of hyperbola, brnaches of hyperbola lie in acute angles formed by asymptotes.
Therefore, eccentricity of hyperbola is `e=sec.(theta)/(2)=sec.(pi)/(6)=(2)/(sqrt3).`