Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The equation of the hyperbola whose foci are  and eccent

The equation of the hyperbola whose foci are \mathrm{(6,5),(-4,5)} and eccentricity \frac{5}{4} is 
 

Option: 1

\mathrm{\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1}


Option: 2

\mathrm{\frac{x^2}{16}-\frac{y^2}{9}=1}


Option: 3

\mathrm{\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=-1}


Option: 4

None of these


Answers (1)

best_answer

\mathrm{\mathrm{S}_1 \equiv(0,5) ; \mathrm{S}_2 \equiv(-4,3), \mathrm{e}=5 / 4}
\mathrm{\mathrm{S}_1 \mathrm{~S}_2=10 \Rightarrow 2 \mathrm{ae}=10 \Rightarrow \mathrm{a}=4}
and \mathrm{\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=16\left(\frac{25}{16}-1\right)=9}
Centre of the hyperbola is (1,5)
Equation of required hyperbola is\mathrm{\frac{(x-1)^2}{16}-\frac{(y-5)^2}{9}=1}

Posted by

Divya Prakash Singh

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

Will there be any changes in the JEE Main 2026 syllabus? subject-wise high-weightage topics
anam-khan

JEE Main 2026 Registration LIVE: NTA JEE application dates soon; apply on jeemain.nta.nic.in
anam-khan

JEE Main 2026: At IIT Bombay, CSE remains top choice; closing rank falls by 2 points

Latest Question