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# The eccentric angles of extremities of a focal chord (other than Major axis) of an ellipse x2a2+y2b2=1 are θ1 and θ2. If the eccentricity of the ellipse are e1 and e2 for the conditions a>b and b>a respectively, then cos2(θ1−θ22)(1e21+1e22) is

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Solution

## Let the coordinates of end points of the chord be P(acosθ1,bsinθ1) and Q(acosθ2,bsinθ2) Equation the Focal chord is, y−bsinθ2x−acosθ2=bsinθ1−bsinθ2acosθ1−acosθ2⇒y−bsinθ2x−acosθ2=−bacos(θ1+θ22)sin(θ1+θ22)⋯(1) When a>b, Focus will be (ae1,0), putting this point in the equation of the focal chord, −bsinθ2ae1−acosθ2=−bacos(θ1+θ22)sin(θ1+θ22)⇒e1cos(θ1+θ22) =cosθ2cos(θ1+θ22)+sinθ2sin(θ1+θ22)⇒e1=cos(θ1−θ22)cos(θ1+θ22) When a<b, Focus will be (0,be2), putting this point in the equation of the focal chord, ⇒be2−bsinθ2−acosθ2=−bacos(θ1+θ22)sin(θ1+θ22)⇒e2=cos(θ1−θ22)sin(θ1+θ22) Therefore, 1e21+1e22=1cos2(θ1−θ22)⇒cos2(θ1−θ22)(1e21+1e22)=1

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