Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A hyperbola whose centre is (0, 0) and passes through (4, 2) and its transverse axis is along x-axis and length of transverse axis is 4, then eccentricity is-

Answers (1)

best_answer

Eccentricities of Hyperbola -

\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}= 1

- wherein

e_{1}\, and \, e_{2}  are eccentricities of the hyperbola and its conjugate.

Now,

Standard equation of the hyperbola is 

\frac{x^2}{a^2} - \frac{y^2}{b^2} =1

length of transverse axis is 4 So,

2a = 4 \Rightarrow a =2

and it passes through (4,2) So,

\frac{x^2}{4} - \frac{y^2}{b^2} =1 

\\4 - \frac{4}{b^2} = 1\\ b^2 = \frac{4}{3}

the eccentricity of a hyperbola : 

e = \sqrt{1 + \frac{b^2}{a^2}}

e = \frac{2}{\sqrt3}

 

Posted by

Pankaj Sanodiya

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 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow
anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern
anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question