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 : $49y^2 – 16x^2 = 784$

$\begin {array} {1 1} (A)\;Foci : (\pm \sqrt{65}, 0), Vertices : (\pm 4, 0), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \\ (B)\;Foci : (0, \pm \sqrt{65}), Vertices : (0, \pm 4), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \\ (C)\;Foci : ( \sqrt{65}, 0), Vertices : ( 4, 0), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \\ (D)\;Foci : (0, - \sqrt{65}), Vertices : (0, - 4), e = \sqrt{65} / 4, Length \: of \: the \: latus \: rectum : 49 / 2 \end {array}$

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 $ 49y^2-16x^2=784$
Dividing throughout by 784 we get,
$ \large\frac{y^2}{16}$$-\large\frac{x^2}{49}$$=1$
Comparing this with the standard equation of hyperbola
$ \large\frac{y^2}{a^2}$$-\large\frac{x^2}{b^2}$$=1$ we get,
$a^2=16 \Rightarrow a = \pm 4$
$ b^2=49 \Rightarrow b \pm 7$
$ \therefore c^2 = a^2+b^2$
(i.e) $c^2 = 16+49 = 65$
$ \therefore c = \sqrt{65}$
Step 2 :
Hence coordinates of foci are : $ ( 0, \pm \sqrt{65} )$
Coordinates of vertices are : $( 0, \pm 4 )$
eccentricity is $ \large\frac{c}{a}$$ = \large\frac{\sqrt{65}}{4}$
Length of the latus rectum is $ \large\frac{2b^2}{a}$
$ = \large\frac{2 \times 49}{4}$$ = \large\frac{49}{2}$
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
...