If the chord through the points whose eccentric angles are α and β on the ellipse x2a2+y2b2=1 passes through the focus (ae,0), then the value of tan α2 tan β2=
A
e+1e−1
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B
e−1e+1
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C
e+1e−2
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D
None
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Solution
The correct option is De−1e+1 Equation of chord to the ellipse with eccentric angle α and β is given by, xacosα+β2+ybsinα+β2=cosα−β2 Given it passes through (ae,0) ⇒ecosα+β2=cosα−β2 ⇒e(cosα2cosβ2−sinα2sinβ2)=cosα2cosβ2+sinα2sinβ2 ⇒e(1−tanα2tanβ2)=1+tanα2tanβ2 ∴tanα2tanβ2=e−1e+1