Question

An ellipse is inscribed in a reactangle and the angle between the diagonals of the reactangle is `tan^(-1)(2sqrt(2))`, then find the ecentricity of the ellipse

Answer 1

Correct Answer - `(1)/(sqrt(2))`

From the figure, we have
`tan.(theta)/(2)=(b)/(a)rArrtan^(2).(theta)/(2)=(b^(2))/(a^(2))`
`rArr(1-costheta)/(1+cos theta)=(b^(2))/(a^(2))`
`rArr(1-(1)/(3))/(1+(1)/(3))=(b^(2))/(a^(2))rArr(b^(2))/(a^(2))=(1)/(2)`
`:. e=sqrt(1-(b^(2))/a^(2))=(1)/(sqrt(2))`