logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  AIMS
0 votes

The foci of the ellipse $\large\frac{x^2}{16}+\frac{y^2}{b^2}$$=1$ and the hyperbola $\large\frac{x^2}{144}-\frac{y^2}{81} =\frac{1}{25}$ coincide. Then the value of $b^2$ is

$\begin{array}{1 1}(A)\;1 \\(B)\;5 \\(C)\;7 \\ (D)\;9 \end{array} $
Can you answer this question?
 
 

1 Answer

0 votes
Then the value of $b^2$ is 7
Hence C is the correct answer.
answered Apr 11, 2014 by meena.p
 

Related questions

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