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  • From a point P(1, 2) pair of tangent’s are drawn to a hyperbola ‘H’ where the two tangents touch different arms of hyperbola. Equation of asymptotes of hyperbola H are<img alt="\s

From a point P(1, 2) pair of tangent’s are drawn to a hyperbola ‘H’ where the two tangents touch different arms of hyperbola. Equation of asymptotes of hyperbola H are\sqrt 3 x - y + 5 = 0 \: \: and \: \: \sqrt 3 x + y - 1 = 0 then eccentricity of ‘H’ is

Option: 1

2 


Option: 2

2 \sqrt 3


Option: 3

\sqrt 2


Option: 4

\sqrt 3


Answers (1)

best_answer

 

Equation of angle bisectors -

\frac{a_{1}x+b_{1}y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}} =\frac{\pm a_{2}x+b_{2}y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}

 

- wherein

Angle bisectors of the lines a_{1}x+b_{1}y+c_{1}=0   and a_{2}x+b_{2}y+c_{2}=0

 

  

Acute and obtuse angle bisectors -

If  \left | tan\Theta \right |<1,  bisector is acute angle bisector. If  \left | tan\Theta \right |>1,  bisector is obtuse angle bisector.

 

- wherein

If \Theta is the angle between one of the lines and one of bisectors.

 

 

 

origin lies in acute angle of asymptotes

P(1, 2) lies in obtuse angle of asymptotes

 acute angle between the asymptotes is 

\pi /3 \\\\ \therefore e = \sec \theta /2 = \sec \pi /6 = 2 \sqrt 3

Posted by

Ritika Kankaria

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