Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A tangent drawn to the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E

A tangent drawn to the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 } at \mathrm{ \quad P\left(\frac{\pi}{6}\right)}  forms a triangle of area \mathrm{3 a^2 } sq. units with coordinate axes. Eccentricity of the hyperbola is equal to

Option: 1

\sqrt{17}


Option: 2

\sqrt{21}


Option: 3

4


Option: 4

\sqrt{6}


Answers (1)

best_answer

Any arbitrary tangent is \mathrm{\frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1}

If it meets the x-axis and y-axis at the points A and B respectively,

\mathrm{\text { then } A \equiv(a \cos \theta, 0), B \equiv(0, b \cot \theta)}

\mathrm{\begin{aligned} & \Delta_{O A B}=\frac{1}{2} a b \frac{\cos ^2 \theta}{\sin \theta}=\frac{a b}{2} \cdot \frac{3}{4} \cdot \frac{2}{1}=\frac{3 a b}{4}=3 a^2 \quad \text { (given) } \\ & \Rightarrow \quad b=4 a \Rightarrow 16 a^2=a^2\left(e^2-1\right) \Rightarrow e=\sqrt{17} \end{aligned}}

 

Posted by

shivangi.shekhar

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

75% criteria for JEE Main 2025 not removed; eligibility criteria for admission to IITs
anam-khan

BTech Admissions 2025 LIVE: AEEE phase 1 exam date out, JEE Mains registration ongoing; BITSAT, VITEEE updates
anam-khan

JEE Main 2025 Eligibility Criteria: What is state code of eligibility? Here’s list of qualifying exams

Latest Question