1.

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»


Solution :
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))`


Discussion

No Comment Found

Related InterviewSolutions