Question

if the length of latus rectum of an ellipse is 1#2 of its major axis then find its eccentricity

Answer 1

e
$e = \sqrt{3} \div 2$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Eccentricity=\frac{1}{\sqrt{2}}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{: \implies {Latus \: rectum} = half\:of\:major\:axis} \\ \\ \red{ \underline \bold{To \: Find: }} \\ \tt{: \implies Eccentricity(e) =?}$

• According to given question :

$\tt{: \implies latus \: rectum=half\:of\:major\:axis} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies \frac{2 {b}^{2} }{a} =\frac{2a}{2}} \\ \\ \tt{: \implies a = \frac{2 {b}^{2} }{a} } \\ \\ \tt{: \implies {a}^{2} = 2{b}^{2} } - - - - - (1) \\ \\ \bold{As \: we \: know \:that} \\ \tt{: \implies {b}^{2} = {a}^{2} ( 1-{e}^{2} )} \\ \\ \text{Putting \: value \: of \: {b}}^{2}\\ \tt{: \implies {b}^{2} = {2b}^{2} ( 1-{e}^{2} )} \\ \\ \tt{: \implies {b}^{2} = 2{b}^{2} ( 1-{e}^{2} )} \\ \\ \tt{: \implies 1 =2-2e^{2}} \\ \\ \tt{: \implies {e}^{2} = \frac{1}{2}} \\ \\ \green{\tt{: \implies e = \frac{1}{\sqrt{2}}}}$