Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt 2$. Its equation is

$\begin{array}{1 1}(A)\;x^2-y^2=32 \\(B)\;\frac{x^2}{4} -\frac{y^2}{9}=1 \\(C)\;2x^2-3y^2=7 \\(D)\; none\;of\;these \end{array}$

Can you answer this question?

1 Answer

0 votes
$ae=8\qquad e=\sqrt 2$
$a=4 \sqrt 2$
$\qquad= 32 (2-1)$
Hence the hyperbola is $x^2-y^2=32$
Hence A is the correct answer.
answered Apr 21, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App