If the length of latus rectum of a horizontal ellipse x2tan2α+y2sec2α=1(α≠π2) is 12, then the possible value(s) of α in (0,π) is/are
A
π12
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B
π6
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C
5π12
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D
7π12
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Solution
The correct option is C5π12 x2tan2α+y2sec2α=1 ⇒x2cot2α+y2cos2α=1 ∵cos2α=cot2α(1−e2) ⇒sin2α=(1−e2) ⇒e2=cos2α(α≠90∘) ⇒e=cosα
∵ Latus rectum =12=2b2a ⇒cotα=4cos2α ⇒1sinα=4cosα ⇒sin2α=12 ⇒2α=nπ+(−1)nπ6 ⇒α=nπ2+(−1)nπ12
Since, we have to take α∈(0,π)
For n=0 ⇒α=π12
and for n=1 ⇒α=π2−π12=5π12