PQ and QR are two focal chords of an ellipse and the eccentric angles of P,Q,R are `2alpha, 2beta, 2 gamma`, respectively then `tan beta gamma` is equal to

Options

A. `cot alpha`
B. `cot^(2)alpha`
C. `2 cot alpha`
D. None of these

Answer

B

Explanation

`(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1`
`P(a cos 2alpha, b sin 2 alpha), Q (a cos 2 beta, b sin 2 beta)`
`R(a cos 2 gamma,b sin 2 gamma)`
Equation of chord PQ is
`(x)/(a) cos (alpha + beta) + (y)/(b) sin (alpha + beta) = cos (alpha - beta)`
PQ passes through the focus (ae,0)
`:. e = (cos(alpha-beta))/(cos(alpha+beta))`
`:. (cos(alpha-beta))/(cos(alpha+beta)) =-(cos(alpha-gamma))/(cos(alpha+gamma))`
Apply componendo and dividendo, we get
`(cos(alpha+beta)+cos(alpha-beta))/(cos(alpha+beta)-cos(alpha-beta))=(cos(alpha+gamma)-cos(alpha-gamma))/(cos(alpha+gamma)+cos(alpha-gamma))`
`(2 cos alpha cos beta)/(2 sin alpha sin beta) = (2 sin alpha sin gamma)/(2 cos alpha cos gamma)`
`tan beta tan gamma = cot^(2) alpha`