Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The distance between the foci of a hyperbola is 16 and its eccentricity is . Its equation is A. x^2 – y^2 = 32 B. x^2/4 - y^2/9=1 C. 2x – 3y^2 = 7 D. none of these

The distance between the foci of a hyperbola is 16 and its eccentricity is \sqrt{2}. Its equation is
A. x2 – y2 = 32
B. \frac{x^{2}}{4}-\frac{y^{2}}{9}=1
C. 2x – 3y2 = 7
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

Given that Foci=(±ae,0)

Distance between foci=2ae=16

 2*a*\sqrt{2}=16

a=4\sqrt{2}

b^{2}=a^{2}(e^{2}-1)=(4\sqrt{2})^{2}((\sqrt{2})^{2}-1)

=32(2-1)=32  

 Hence, equation is \frac{x^{2}}{32}-\frac{y^{2}}{32}=1

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question