Question

If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

• A. $\frac{\cos\,\alpha+\cos\,\beta}{\cos(\alpha-\beta)}$
• B. $\frac{\sin\,\alpha-\sin\,\beta}{\sin(\alpha-\beta)}$
• C. $\frac{\cos\,\alpha-\cos\,\beta}{\cos(\alpha-\beta)}$
• D. $\frac{\sin\,\alpha+\sin\,\beta}{\sin(\alpha-\beta)}$

Answer 1

Answer: (D)

The equation of the chord joining the points $P\left(\alpha\right), Q\left(\beta\right)$ is given by
$\frac{x}{a} cos\frac{ \alpha+\beta}{2} + \frac{y}{b} sin \frac{ \alpha +\beta }{2} = cos\frac{ \alpha -\beta }{2}$
It passes thro' focus $\left(ae, 0\right)$, then
$\frac{ae}{a} cos \frac{ \alpha +\beta }{2} = cos\frac{ \alpha -\beta }{2}$
$\Rightarrow e\, cos\frac{ \alpha +\beta }{2} = cos \frac{ \alpha -\beta }{2} $
$\Rightarrow \frac{e}{1} = \frac{cos\frac{ \alpha -\beta }{2}}{cos\frac{ \alpha +\beta }{2}} $
$\Rightarrow e =\frac{ 2\,sin \frac{ \alpha +\beta }{2} cos \frac{ \alpha -\beta }{2}}{2\,sin \frac{ \alpha +\beta }{2} cos\frac{ \alpha +\beta }{2}} $
$= \frac{sin \alpha +sin\beta}{sin \left(\alpha+\beta\right)}$