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Global Maxima

The eccentric...

Question

The eccentric angles of the extremities of latus rectum of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ are given by?

{tan}^{- 1} (\pm \frac{a e}{b})

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{tan}^{- 1} (\pm \frac{b e}{a})

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{tan}^{- 1} (\pm \frac{b}{a e})

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{tan}^{- 1} (\pm \frac{a}{b e})

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Solution

The correct option is C ${tan}^{- 1} (\pm \frac{b}{a e})$
Correct option is (C)

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Similar questions

α

β

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

(a e, 0),

\frac{α}{2}

\frac{β}{2} =

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} + y^{2} = a b

{tan}^{- 1} \frac{(b - a)}{\sqrt{a b}}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} + y^{2} = a b

⎛ ⎝ \sqrt{\frac{a^{2} b}{a + b}}, \sqrt{\frac{a b^{2}}{a + b}} ⎞ ⎠

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a > b

e

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; a > b

e

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

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