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  • Equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0) is A. x^2/4 - y^2/5=4/9 B. x^2/9 - y^2/9=4/9 C. x^2/4 - y^2/9=1 D. None of these

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

A. \frac{x^{2}}{4}-\frac{y^{2}}{5}=\frac{4}{9}

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

C. \frac{x^{2}}{4}-\frac{y^{2}}{9}=1

D. None of these
 

Answers (1)

Option (A) is correct.

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

 e=3/2 

 Foci=(±ae),0=(±2,0)

  So, after comparing the equations, ae=2  

a*3/2=2  

 a=4/3 

  b2=a2(e2-1)   

 b2=(4/3)2((3/2)2-1) 

=(16/9)(9/4-1)

=16/9*5/4=20/9   

  Equation of hyperbola is \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1

=\frac{x^{2}}{\left (\frac{4}{3} \right )^{2}}-\frac{y^{2}}{\frac{20}{9}}=1

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

  Hence, \frac{x^{2}}{16}-\frac{y^{2}}{{20}}=\frac{1}{9} is the required equation. 

 

Posted by

infoexpert22

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