If α and β are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

Question

If α and β are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is (A) ( cos α+ cos β/ cos(α-β)) (B) ( sin α- sin β/ sin(α-β)) (C) ( cos α- cos β/ cos(α-β)) (D) ( sin α+ sin β/ sin(α-β))

Tardigrade Admin · Accepted Answer

The equation of the chord joining the points

P (α), Q (β)

is given by

\frac{x}{a} cos \frac{α + β}{2} + \frac{y}{b} s in \frac{α + β}{2} = cos \frac{α - β}{2}

It passes thro' focus

(a e, 0)

, then

\frac{a e}{a} cos \frac{α + β}{2} = cos \frac{α - β}{2}

\Rightarrow e cos \frac{α + β}{2} = cos \frac{α - β}{2}

\Rightarrow \frac{e}{1} = \frac{cos \frac{α - β}{2}}{cos \frac{α + β}{2}}

\Rightarrow e = \frac{2 s in \frac{α + β}{2} cos \frac{α - β}{2}}{2 s in \frac{α + β}{2} cos \frac{α + β}{2}}

= \frac{s in α + s in β}{s in ( α + β )}

Q. If $α$ and $β$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

$\frac{c o s α + c o s β}{c o s ( α - β )}$

$\frac{s i n α - s i n β}{s i n ( α - β )}$

$\frac{c o s α - c o s β}{c o s ( α - β )}$

$\frac{s i n α + s i n β}{s i n ( α - β )}$

Q. If α and β are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

cos(α−β)cosα+cosβ​

sin(α−β)sinα−sinβ​

cos(α−β)cosα−cosβ​

sin(α−β)sinα+sinβ​

Q. If $α$ and $β$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

$\frac{c o s α + c o s β}{c o s ( α - β )}$

$\frac{s i n α - s i n β}{s i n ( α - β )}$

$\frac{c o s α - c o s β}{c o s ( α - β )}$

$\frac{s i n α + s i n β}{s i n ( α - β )}$