Question

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

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

Let $a,b$ are the length of semi-major axis and semi-minor axis respectively, then $(a\cos \alpha ,b\sin \alpha ),(a\cos \beta ,b\sin \beta ),(ae,0)$ are collinear.

$\Rightarrow \frac{b(\sin \beta -\sin \alpha )}{a(\cos \beta -\cos \alpha )}=\frac{b\sin \alpha -0}{a\cos \alpha -ae}\Rightarrow (\cos \alpha -e)(\sin \beta -\sin \alpha )=\sin \alpha (\cos \beta -\cos \alpha )\Rightarrow e=\frac{\cos \alpha (\sin \beta -\sin \alpha )-\sin \alpha (\cos \beta -\cos \alpha )}{\sin \beta -\sin \alpha }\Rightarrow e=\frac{\sin (\alpha -\beta )}{\sin \alpha -\sin \beta }\Rightarrow e=\frac{\sin (\alpha -\beta )(\sin \alpha +\sin \beta )}{{\sin }^2\alpha -{\sin }^2\beta }\therefore e=\frac{\sin \alpha +\sin \beta }{\sin (\alpha +\beta )}[\because \sin (A+B\left)\sin \right(A-B)={\sin }^{2}A-{\sin }^{2}B]$