Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).

Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).

Answers (1)

Let the equation of the hyperbola be \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

ae=2

a*\frac{3}{2}=2

a=\frac{4}{3}

b2=a2(e2-1)

b^{2}=\frac{16}{9}*\left ( \frac{9}{4}-1 \right )=\frac{20}{9}

\frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{\frac{20}{9}}=1

\frac{x^{2}}{16}-\frac{y^{2}}{20}=\frac{1}{9}

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question