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

If latus rectum of ellipse `x^2tan^2phi+y^2sec^2phi=1` is 1/2, then th

1.	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
Answer» `x^2/cos^2phi+y^2/cos^2phi=1` `a^2=cot^2phi` `b^2=cos^2phi` `1/2=(2b^2)/a=(2cos^2phi)/cotphi` `1/2=(2cos^2phisinphi)/cosphi` `1/2=sin^2phi` `2phi=pi/6` `phi=pi/12` `sin(pi-2phi)=1/2` `pi-2phi=pi/6` `pi-pi/6=2phi` `2phi=5/6pi` `phi=5/12pi` option d is correct.

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