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If eccentric angle of a point lying in t...

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

A

`tan^(-1) sqrt((a)/(b))`

B

`tan^(-1)sqrt((b)/(a))`

C

`(pi)/(4)`

D

`(pi)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
According to question `tan phi = (b)/(a) tan theta`
`tan (theta - phi) = (tan theta - tan phi)/( 1+ tan theta tan phi)`
`=(tan theta-(b)/(a)tan theta)/(1+tan theta((b)/(a)tan theta))`
`= (a-b)/(a cot theta + b tan theta)`
This will be maximum if `tan^(2) theta = (a)/(b) rArr tan theta = sqrt((a)/(b))`
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