logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

The ecentricity of an ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ whose latus rectum is half to its major axis is e.Find the value of e?

$\begin{array}{1 1}(a)\;\large\frac{1}{\sqrt 2}\\(b)\;\large\frac{1}{\sqrt 3}\\(c)\;\large\frac{1}{\sqrt 5}\\(d)\;\text{None of these}\end{array}$

1 Answer

Latus rectum =$\large\frac{2b^2}{a}$
According to the question
$\large\frac{2b^2}{a}=\frac{1}{2}$$\times 2a$
$\large\frac{b^2}{a}=\frac{1}{2}$
We know $b^2=a^2(1-e^2)$
$\large\frac{b^2}{a}=$$1-e^2$
$\large\frac{1}{2}=$$1-e^2$
$\large\frac{1}{2}$$-1=e^2$
$e=\large\frac{1}{\sqrt 2}$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
edited Sep 23, 2014 by sharmaaparna1
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X