If `omega` is one of the angles between the normals to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1' at the point whose eccentric angles are `theta` and `pi/2+theta` , then prove that `(2cotomega)/(sin2theta)=(e^2)/(sqrt(1-e^2))`

Options

A. `(e^(2))/(sqrt(1-e^(2)))`
B. `(e^(2))/(sqrt(1+e^(2)))`
C. `(e^(2))/(1-e^(2))`
D. `(e^(2))/(1+e^(2))`

Answer

A

Explanation

The equations of the normals to the ellipse `(x^(2))/(a^(2)) + (y^(2))/(b^(2)) =1` at the points whose eccentric angles are `theta` and `(pi)/(2) + theta` are
`ax sec theta - by cosec theta = a^(2) -b^(2)` and `-ax cosec theta - by sec theta = a^(2) -b^(2)` respectively.
Since `omega` is the angle between these two normals. Therefore,
`tan omega=|((a)/(b)tan theta+(a)/(b)cot theta)/(1-(a^(2))/(b^(2)))|`
`rArr tan omega = |(ab(tan theta + cot theta))/(b^(2)-a^(2))|`
`rArr tan omega = |(2ab)/((sin 2 theta)(b^(2)-a^(2)))|`
`rArr tan omega = (2ab)/((a^(2)-b^(2))sin 2 theta)`
`rArr tan omega = (2a^(2)sqrt(1-e^(2)))/(a^(2)e^(2)sin2 theta)`
`rArr (2 cot omega)/(sin 2 theta) = (e^(2))/(sqrt(1-e^(2)))`