Question

If $\alpha$ and $\beta$ are the eccentric angles of the extremeties of the chord of the standard horizontal ellipse, and passes through the focus $(ae,0)$, then eccentricity of the ellipse is

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

Answer 1

The correct option is A $\frac{\sin \alpha +\sin \beta }{\sin (\alpha +\beta )}$

Given: Chord passing through the focus $(ae,0)$ and eccentric angles of the extremeties of the chord are $\alpha$ and $\beta$.

To Find: Eccentricity of the ellipse

Step-$1:$ Consider the equation of the focal chord

Step-$2:$ Use the coordinates of the foci to find the value of eccentricity.

If $\alpha$ and $\beta$ are the extremeties of the focal chord $d=ae$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{e-1}{e+1}$

$\Rightarrow \frac{e-1}{e+1}=\frac{\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}{\cos \frac{\alpha }{2}\cos \frac{\beta }{2}}$

apply componendo and dividendo.

$\Rightarrow \frac{e-1+e+1}{e-1-e-1}=\frac{\sin \frac{\alpha }{2}\sin \frac{\beta }{2}+\cos \frac{\alpha }{2}\cos \frac{\beta }{2}}{\sin \frac{\alpha }{2}\sin \frac{\beta }{2}-\cos \frac{\alpha }{2}\cos \frac{\beta }{2}}$

$\Rightarrow \frac{2e}{-2}=\frac{\cos (\frac{\alpha -\beta }{2})}{-\cos (\frac{\alpha +\beta }{2})}$

$\Rightarrow e=\frac{\cos (\frac{\alpha -\beta }{2})}{\cos (\frac{\alpha +\beta }{2})}$

Multiply and divide $2\sin (\frac{\alpha +\beta }{2})$

$\Rightarrow e=\frac{2\sin (\frac{\alpha +\beta }{2})\cos (\frac{\alpha -\beta }{2})}{2\sin (\frac{\alpha +\beta }{2})\cos (\frac{\alpha +\beta }{2})}$

$e=\frac{\sin \alpha +\sin \beta }{\sin (\alpha +\beta )}$