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# hey how to find eccentric angles at the end points of latus rectum of the ellipse x sq by 4 PLUS y sq = 1 ?

Jitender Singh IIT Delhi
7 years ago
Ans: 30
Sol:
Ellipse:
$\frac{x^{2}}{4}+y^{2}=1$
Let the end of the latus rectum on the ellipse be P & Q. Then
$P(acos\theta , bsin\theta )$
$Q(acos\theta , -bsin\theta )$
Since latus rectum passes through focus, we have
$acos\theta =ae$
$cos\theta =e$
$e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1- \frac{1}{4} = \frac{3}{4}$
$e= \frac{\sqrt{3}}{2}$
$cos\theta = \frac{\sqrt{3}}{2}$
$\theta = cos^{-1}\frac{\sqrt{3}}{2}$
$\theta = 30$
Thanks & Regards
Jitender Singh
IIT Delhi