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Question

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(θ1θ22)(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,
ybsinθ2xacosθ2=bsinθ1bsinθ2acosθ1acosθ2ybsinθ2xacosθ2=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,
bsinθ2ae1acosθ2=bacos(θ1+θ22)sin(θ1+θ22)e1cos(θ1+θ22) =cosθ2cos(θ1+θ22)+sinθ2sin(θ1+θ22)e1=cos(θ1θ22)cos(θ1+θ22)

When a<b,
Focus will be (0,be2), putting this point in the equation of the focal chord,
be2bsinθ2acosθ2=bacos(θ1+θ22)sin(θ1+θ22)e2=cos(θ1θ22)sin(θ1+θ22)

Therefore,
1e21+1e22=1cos2(θ1θ22)cos2(θ1θ22)(1e21+1e22)=1

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