If `alpha and beta` are eccentric angles of the ends of a focal chord of the ellipse `x^2/a^2 + y^2/b^2 =1`, then `tan alpha/2 .tan beta/2` is (A) `(1-e)/(1+e)` (B) `(e+1)/(e-1)` (C) `(e-1)/(e+1)` (D) none of these

Options

A. `(1-e)/(1+e)`
B. `(e-1)/(e+1)`
C. `(e+1)/(e-1)`
D. none of these

Answer

B

Explanation

the coordinates of the end points of the focal chord are ( a cos `alpha , b sin alpha ) and ( a cos beta , bsin beta)` therefore , the equation of the focal chgord is
`(x)/(a)cos ((alpha + beta)/(2))+ (y)/( b) sin ((alpha+ beta)/(2)) = cos (( alpha - beta)/(2))`
this passes through ( ae,0)
`therefore (ae)/(0) cos ((alpha + beta)/(2))= cos ((alpha - beta)/(2))`
`implies ( cos ((alpha +beta)/(2)))/(cos ((alpha +beta)/(2)))=(1)/(e)`
`implies ( cos ((alpha + beta)/( 2))- cos( ( alpha - beta)/(2)))/(cos ((alpha + beta)/(2))+ cos ((alpha- beta)/(2)))=(1-e)/(1+e)`
`implies -(2 sin ""(alpha)/(2) sin ""( beta)/(2))/(2 cos "" ( alpha)/(2) cos ( beta)/(2) )=(1-e)/(1+e)`
`implies - tan ""( alpha ) /( 2) tan "" ( beta)/(2) =(1-e)/(1+e)implies tan ""(alpha)/(2) tan""(beta)/(2)=(e-1)/( e+1)`