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 of an ellipse that satisfies the following conditions: Ends of Major axis $\;(\pm3,0)$, Ends of Minor Axis $\;(0, \pm 2)$

$\begin{array}{1 1}\frac{x^2}{9}+\frac{y^2}{4}=1 \\ \frac{x^2}{9}+\frac{y^2}{8}=1 \\ \frac{x^2}{4}+\frac{y^2}{9}=1 \\ \frac{x^2}{16}+\frac{y^2}{36}=1 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Given an ellipse as follows:
  • http://clay6.com/mpaimg/Toolbar_7.png
  • The equation of the ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • $c= \sqrt {a^2 - b^2}$
  • Given the major and minor axis, we can substitute in the above equation and arrive at the equation.
Given Ends of Major axis $\;(\pm3,0)$, Ends of Minor Axis $\;(0, \pm 2)$ we get $a = 3$ and $b = 2$.
Here the major axis is the x-axis.
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{3^2}$$+\large\frac{y^2}{2^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{9}$$+\large\frac{y^2}{4}$$=1$
answered Apr 2, 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
...