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If latus rectum of ellipse x^2tan^2phi+y...

If latus rectum of ellipse `x^2tan^2phi+y^2sec^2phi=1` is 1/2, then the value of `phi` where `phi in (0,pi)` is

Text Solution

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`x^2/cos^2phi+y^2/cos^2phi=1`
`a^2=cot^2phi`
`b^2=cos^2phi`
`1/2=(2b^2)/a=(2*cos^2phi)/cotphi`
`1/2=(2cos^2phi*sinphi)/cosphi`
`1/2=sin^2phi`
`2phi=pi/6`
`phi=pi/12`
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