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 coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas : $9y^2–4x^2= 36$

$\begin {array} {1 1} (A)\;Foci : (0, \pm \sqrt {13}/2 ), Vertices : (0, \pm 2), e = \sqrt{13} / 4, Length \: of \: the \: latus \: rectum : 9 \\ (B)\;Foci : ( \pm \sqrt {13}/2, 0 ), Vertices : ( \pm 2, 0), e = \sqrt{13} / 4, Length \: of \: the \: latus \: rectum : 9 \\ (C)\;Foci : (0, \pm \sqrt {13} ), Vertices : (0, \pm 2), e = \sqrt{13} / 4, Length \: of \: the \: latus \: rectum : 9 \\ (D)\;Foci : ( \pm \sqrt {13}, 0 ), Vertices : (0, \pm 2), e = \sqrt{13}/ 4, Length \: of \: the \: latus \: rectum : 9 \end {array}$

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 :
The given equation is $ 9y^2-4x^2=36$
Dividing throughout by 36 we get,
$ \large\frac{y^2}{4}$$- \large\frac{x^2}{9}$$=1$
Comparing this with the standard equation of the hyperbola
$ \large\frac{y^2}{a^2}$$-\large\frac{x^2}{b^2}$$=1$ we get,
$ a^2 = 4 $ and $ b^2=9$
$ \therefore c^2 = 4+9=13$
$ \Rightarrow c = \pm 13$
Step 2 :
Hence the coordinates of foci are $(0, \pm 13)$
The coordinates of vertices are $(0, \pm 2)$
Eccentricity $ e = \large\frac{c}{a}$$= \large\frac{\sqrt{13}}{2}$
Length of latus rectum $ = \large\frac{2b^2}{a}$$ = \large\frac{2 \times 9}{2}$$ = 9$
answered Jul 19, 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
...