If α and β are the eccentric angles of points of contract of tangents drawn from (3,2) to the ellipse x29+y24=1 then |tan(α−β2)|=
A
1
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B
12
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C
5
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D
6
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Solution
The correct option is A1 Point of intersection of the tangents at α,βin(acosα+β2cosα−β2,bsinα+β2cosα−β2) Which is given as (3, 2) ⇒cos2(α−β2)=12 ⇒|tan(α−β2)|=1