Ask Questions, Get Answers

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

Find the value of p for which $e=\large\frac{1}{2}$ where ellipse is $\large\frac{x^2}{4}+\frac{y^2}{p^2}$$=1$?

$\begin{array}{1 1}(a)\;p=\sqrt 3,\large\frac{4}{\sqrt 3}\\(b)\;p=\sqrt 5,\large\frac{4}{\sqrt 5}\\(c)\;p=\sqrt 7,\large\frac{4}{\sqrt 7}\\(d)\;\text{None of these}\end{array}$

1 Answer

Two cases are possible
(i) 4 > p
$p=\sqrt 3$
(ii) 4 < p
$p=\large\frac{4}{\sqrt 3}$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v

Related questions