If `alphaa n dbeta` are the eccentric angles of the extremities of a focal chord of an ellipse, then prove that the eccentricity of the ellipse is `(sinalpha+sinbeta)/("sin"(alphabeta+))`

Options

A. `(cos alpha + cos beta)/(cos (alpha - beta))`
B. `(sin alpha - sin beta)/(sin (alpha - beta))`
C. `(cos alpha - cos beta)/(cos (alpha - beta))`
D. `(sin alpha + sin beta)/(sin(alpha + beta))`

Answer

D

Explanation

The equation fo a chord joining points having eccentric angles `alpha` and `beta` is given by
`x/a cos ((alpha + beta)/(2)) + y/b sin ((alpha + beta)/(2)) = cos ((alpha - beta)/(2))`
If the passes through (ae, 0) then
`e cos ((alpha + beta)/(2)) = cos ((alpha - beta)/(2))`
`rArr e = (cos((alpha - beta)/(2))/(cos ((alpha + beta)/(2))`
`rArr e = (2 sin ((alpha + beta)/(2))cos((alpha - beta)/(2))/2 sin ((alpha + beta)/(2))cos((alpha + beta)/(2)) = (sin alpha + sin beta)/(sin (alpha + beta))`