Question

If eccentric angle of a point lying in the first quadrant on the ellipse `x^2/a^2 + y^2/b^2 = 1` be `theta` and the line joining the centre to that point makes an angle `phi` with the x-axis, then `theta - phi` will be maximum when `theta` is equal to

Answer 1

`tanphi=(bsintheta)/(acostheta)`
`tanphi=b/atantheta`
`tan(theta-phi)=(tantheta-tanphi)/(1+tanthetatanphi`
=`(tantheta-b/atantheta)/(1+tantheta*b/atantheta`
`=((a-b)tantheta)/(a+btan^2theta)`
`tan(theta-phi)=(a-b)/(acottheta+btantheta)`
`(acottheta+btantheta)/2>=sqrt(acottheta-btantheta)`
`acottheta+btantheta>=2sqrt(ab)`
this is minimum.