Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve it, - Two-dimensional Coordinate Geometry - BITSAT

Equation of the hyperbola with eccentricity 3/2 and foci at (\pm2, 0) is

  • Option 1)

    \frac{x^{2}}{4}-\frac{y^{2}}{9}=\frac{4}{9}

  • Option 2)

    \frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{4}{9}

  • Option 3)

    \frac{x^{2}}{4}-\frac{y^{2}}{9}=1

  • Option 4)

    None of these

 

Answers (1)

best_answer

 

Coordinates of foci -

\left ( \pm \, ae ,o\right )

- wherein

For the Hyperbola

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

 

 ae=2

\Rightarrow a=\frac{4}{3}

b^{2}=a^{2}(e^{2}-1)

\Rightarrow b^{2}=a^{2}(e^{2}-1)

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

=\frac{20}{9}                                    \Rightarrow b=\frac{2}{3} \sqrt{5}

equation \frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{\frac{20}{9}} = 1 \Rightarrow \frac{9x^{2}}{10}-\frac{9y^{2}}{20} = 1

 


Option 1)

\frac{x^{2}}{4}-\frac{y^{2}}{9}=\frac{4}{9}

This solution is incorrect

Option 2)

\frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{4}{9}

This solution is incorrect

Option 3)

\frac{x^{2}}{4}-\frac{y^{2}}{9}=1

This solution is incorrect

Option 4)

None of these

This solution is correct

Posted by

divya.saini

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 application correction facility begins tomorrow; editable fields
anam-khan

BITSAT 2024 registration window closes tomorrow; exam from May 20
anam-khan

BITSAT 2024 registration deadline extended till April 16; session 1 exam from May 19 to 24

Latest Question