The eccentric angle of the end point of the l | vedclass

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Question

(C) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,any point with eccentric angle $	heta$ has coordinates $(a \cos 	heta, b \sin 	heta)$.Th

Vedclass · Accepted Answer

$	an^{-1} \left( \frac{\pm b}{ae} ight)$

Vedclass · Answer

(C) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,any point with eccentric angle $	heta$ has coordinates $(a \cos 	heta, b \sin 	heta)$.The coordinates of the end points of the latus rectum are $(ae, \pm \frac{b^2}{a})$.Equating the coordinates,we have $a \cos 	heta = ae$ and $b \sin 	heta = \pm \frac{b^2}{a}$.From $a \cos 	heta = ae$,we get $\cos 	heta = e$.From $b \sin 	heta = \pm \frac{b^2}{a}$,we get $\sin 	heta = \pm \frac{b}{a}$.Therefore,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\pm b/a}{e} = \pm \frac{b}{ae}$.Thus,$	heta = 	an^{-1} \left( \pm \frac{b}{ae} ight)$.

The eccentric angle of the end point of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

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