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# 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}$$=1? \begin{array}{1 1}(a)\;p=\sqrt 3,\large\frac{4}{\sqrt 3}\\(b)\;p=\sqrt 5,\large\frac{4}{\sqrt 5}\\(c)\;p=\sqrt 7,\large\frac{4}{\sqrt 7}\\(d)\;\text{None of these}\end{array} Can you answer this question? ## 1 Answer 0 votes Two cases are possible (i) 4 > p b^2=a^2(1-e^2) p^2=4(1-e^2) \large\frac{p^2}{4}$$-1=-e^2$
$p^2=4(1-\large\frac{1}{4})$
$p^2=3$
$p=\sqrt 3$
(ii) 4 < p
$a^2=b^2(1-e^2)$
$4=p^2(1-e^2)$
$p^2=\large\frac{4}{3/4}$
$p^2=\large\frac{16}{3}$
$p=\large\frac{4}{\sqrt 3}$
Hence (a) is the correct answer.