logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

Find the equations of the hyperbola satisfying the given conditions : Foci $(0, \pm13)$, the conjugate axis is of length $24$.

$\begin {array} {1 1} (A)\;\large\frac{x^2}{144}-\large\frac{y^2}{25}=1 & \quad (B)\;\large\frac{y^2}{25}-\large\frac{x^2}{144}=1 \\ (C)\;\large\frac{x^2}{144}+\large\frac{y^2}{25}=1 & \quad (D)\;\large\frac{x^2}{25}+\large\frac{y^2}{144}=1 \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Standard equation of a hyperbola where transverse axis is along x - axis is $ \large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2} $$=1$
  • Standard equation of a hyperbola whose transverse axis is along y - axis is $ \large\frac{y^2}{a^2} $$ - \large\frac{x^2}{b^2}$$=1$
  • $c^2=a^2+b^2$, where $c$ is the foci
  • Eccentricity $e = \large\frac{c}{a}$
  • Length of the latus rectum $ = \large\frac{2b^2}{a}$
Step 1 :
It is given that the coordinates of foci are $(0, \pm 13)$ and the length of the conjugate axis is 24.
Here the foci are on the y - axis.
Hence the equation of the hyperbola is of the form $ \large\frac{y^2}{a^2}$$-\large\frac{x^2}{b^2}$$=1$
The length of the conjugate axis is 24.
$ \Rightarrow 2b = 24$
$ \therefore b = 12$
Step 2 :
$a^2+b^2 = c^2 \quad \Rightarrow b^2 = c^2-a^2$
Substituting the values of c and a we get,
$ b^2=169-144$
$b^2=25$
Hence the equation of the hyperbola is
$ \large\frac{y^2}{25}$$ - \large\frac{y^2}{144}$$=1$
answered Jul 21, 2014 by thanvigandhi_1
 

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
...