Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(ab) is tan -1 (+-b/ae) please answer quickly

Question

Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(a>b) is tan -1 (+-b/ae) please answer quickly

Arun · Answer

Dear student Let a cos @ = ae b sin @ = ± b^2/a b/a tan@ = ± b^2 /a^2 e hence tan@ = ± b/ ae hope it helps Regards Arun

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Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(a>b) is tan -1 (+-b/ae) please answer quickly

Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(a>b) is tan^-1(+-b/ae)
please answer quickly

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Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(a>b) is tan -1 (+-b/ae) please answer quickly

Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(a>b) is tan-1(+-b/ae)please answer quickly

1 Answers

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Prove that the eccentric angles of the extremities of latus recta of the ellipse x^2/a^2+y^2/b^2=1(a>b) is tan^-1(+-b/ae)
please answer quickly