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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 toA. `cot alpha`B. `cot^(2)alpha`C. `2 cot alpha`D. None of these |
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Answer» Correct Answer - B `(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` |
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