Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If  and <img alt="\mathrm{e^{\prime}}" src="https://entrancecorner.oncodecogs.com/gif.l

If \mathrm{e} and \mathrm{e^{\prime}} be the eccentricities of a hyperbola and its conjugate, then \mathrm{\frac{1}{\mathrm{e}^2}+\frac{1}{\left(\mathrm{e}^{\prime}\right)^2}} is equal to

Option: 1

\mathrm{0}


Option: 2

\mathrm{1}


Option: 3

\mathrm{2}


Option: 4

\mathrm{\sqrt{2}}


Answers (1)

best_answer

Hyperbolas    \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}= \pm 1} are conjugate to each other.

Let  \mathrm{e}  be the eccentricity of   \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} and  \mathrm{e^{\prime}} be the eccentricity of

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

     \mathrm{e^2=1+\frac{a^2}{b^2}}

     and  \mathrm{\left(e^{\prime}\right)^2=1+\frac{b^2}{a^2}}

     Now  \mathrm{\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1}

Posted by

Sayak

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question