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 hyperbola $\; 49y^2-16x^2=784$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
Given $\; 49y^2-16x^2=784$. Dividing by $784, \rightarrow \large\frac{49}{784}$$y^2-\large\frac{16}{784}$$x^2=1 \rightarrow \large\frac{y^2}{16} $$-\large\frac{x^2}{49}$$=1$
On comparing the given equation with the standard equation of a hyperbola, we get:
$\quad a =4, b = 7$
Since $a^2+b^2 = c^2 \rightarrow c = \sqrt{16+49} = \sqrt{65}$
1. Coordinates of Foci $ = (0, \pm \sqrt{65})$
2. Coordinates of Vertices $= (0,\pm 4)$
3. Eccentricity $e = \large\frac{c}{a} $$= \large\frac{\sqrt{65}}{4}$
4. Latus Rectum $=\large\frac{2b^2}{a}$$ = \large\frac{2\times 49}{4} $$= \large\frac{49}{2}$
answered Apr 8, 2014 by balaji.thirumalai
edited Apr 8, 2014 by balaji.thirumalai
 

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
...