Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main admit card 2025 shows NTA tightening security, Here's how
anam-khan

Engineering aspirant dies by suicide in Kota, fourth case this year
anam-khan

What is non-Aadhaar declaration in JEE Main 2025? Explain mismatch between image, face

Latest Question