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 equation for the ellipse that satisfies the given conditions : Ends of major axis $(\pm 3, 0)$ , ends of minor axis $(0, \pm 2)$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{9}+\large\frac{y^2}{4}=1 & \quad (B)\;\large\frac{x^2}{4}+\large\frac{y^2}{9}=1 \\ (C)\;\large\frac{x^2}{9}-\large\frac{y^2}{4}=1 & \quad (D)\;\large\frac{x^2}{4}-\large\frac{y^2}{9}=1 \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If the major axis is along x - axis then the equation of ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • If the major axis is along y - axis then the equation of ellipse is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
  • $c = \sqrt{a^2+b^2}$ , where $c$ is the foci of the ellipse.
It is given that the ends of the major axis is $(\pm 3, 0)$
Ends of minor axis is $(0, \pm 2)$
Hence the major axis is along x - axis.
Hence the equation of ellipse is of the form : $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
where $a^2=(3)^2=9$ and $ b^2=(2)^2=4$
$ \therefore $ Equation of the ellipse is
$ \large\frac{x^2}{9}$$+\large\frac{y^2}{4}$$=1$
answered Jul 18, 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
...