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Find the eccentric angles of the extremi...

Find the eccentric angles of the extremities of the latus recta of the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`

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The correct Answer is:
`tan^(-1)((b)/(ae)),pi-tan^(-1)((b)/(ae)),2pi-tan^(-)((b)/(ae))`
The coordinates of any points on the ellipse `(x^(2))/(a^(2))=(y^(2))/(b^(2))=1` whose eccentric angle is `theta are (a cos theta, b sin theta)`.
The coordinates of the endpoints of latus reactum are `(ae,+-b^(2)//a)` . Therefore, `a cos theta=ae and b sin theta=+-(b^(2))/(a)`
or `tan theta=+-(b)/(ae)`
Hence, four points of latus rectum have eccentric angles `tan^(-1)(b//ae),pi-tan^(-1)(b//ae),pi+tan^(-1)(b//ae), and 2pi-tan^(-1)(b//ae)` in the first, second, third, and fourth quandrants, respecitvely.
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