Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The eccentricity of the hyperbola which passes through and <img

The eccentricity of the hyperbola which passes through \left ( 3,0 \right ) and \left ( 3\sqrt{2},2 \right ) is

Option: 1

\sqrt{(13)}


Option: 2

\frac{\sqrt{13}}{3}


Option: 3

\sqrt{\frac{13}{4}}


Option: 4

None of these


Answers (1)

best_answer

Let equation of hyperbola is \mathrm{x^2 / a^2-y^2 / b^2=1}.  Point (3, 0) lies on hyperbola
So, \mathrm{\frac{(3)^2}{a^2}-\frac{0}{b^2}=1 \quad \text { or } \quad \frac{9}{a^2}=1 \text\ { or }\ a^2=9} and point \mathrm{{(3\sqrt{2}, 2)} also lies on hyperbola.  So,
\mathrm{\frac{3(\sqrt{2})^2}{a^2}-\frac{(2)^2}{b^2}=1}
Put \mathrm{a^{2}=9} we get, \mathrm{\frac{18}{9}-\frac{4}{b^2}=1 \quad \text\ { or }\ \ 2-\frac{4}{b^2}=1 \text { or }-\frac{4}{b^2}=1-2 \text { or } \frac{4}{b^2}=1 \text { or } b^2=4}
We know that , \mathrm{b^2=a^2\left(e^2-1\right)} Putting values of \mathrm{a^2} and \mathrm{b^2}
\mathrm{4=9\left(e^2-1\right) \text { or } \quad e^2-1=\frac{4}{9} \text { or } \quad e^2=1+\frac{4}{9} \text { or } \quad e=\sqrt{(1+4 / 9)} \text { or } e=\sqrt{(13) / 9}=\frac{\sqrt{13}}{3}}

Posted by

chirag

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 Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?
anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?
anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question