Consider an ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1(a gt b)`. A hyperbola has its vertices at the extremities of minor axis of the ellipse and the length of major axis of the ellipse is equal to the distance between the foci of hyperbola. Let `e_(1)` and `e_(2)` be the eccentricities of ellipse and hyperbola, respectively. Also, let `A_(1)` be the area of the quadrilateral fored by joining all the foci and `A_(2)` be the area of the quadrilateral formed by all the directries.
The relation between `e_(1)` and `e_(2)` is given by

Options

A. `tan^(-1)((1)/(sqrt(1-e_(1)^(2))))`
B. `tan^(-1)((e_(1))/(sqrt(1-e_(1)^(2))))`
C. `tan^(-1)((1)/(sqrt(e_(2)^(2)-1)))`
D. `tan^(-1)sqrt(1=e_(1)^(2))`

Answer

C

Explanation

We have
`b^(2)=a^(2)(1-e_(1)^(2))`
`"and "2be^(2)=2arArre_(2)=(a)/(b)`
`"So, "(1)/(e_(2)^(2))=1-e_(1)^(2)`
`rArr" "e_(2)^(2)(1-e_(1)^(2))=1`
Tangent at point P `(a cos theta, b sin theta)` on the ellipse is
`(x)/(a) cos theta+(y)/(b) sin theta=1`
It passes through `(0, be_(2))`.
`"So, "e_(2) sin theta=1`
`rArr" "sin theta=(1)/(e_(2))`
`therefore" "theta=tan^(2)((1)/(sqrt(e_(2)^(2)-1)))`
`A_(1)=4xx(1)/(2)xxae_(1)xxbe_(2)=2abe_(1)e_(2)`
`A_(2)=((2a)/(e_(1)))((2b)/(e_(2)))=(4ab)/(e_(1)e_(2))`
`(A_(1))/(A_(2))=(e_(1)^(2)e_(2)^(2))/(2)=2`
`rArr" "e_(1)e_(2)=2`
`"But "e_(2)^(2)(1-e_(1)^(2))=1`
`"So, "e_(2)^(2)-4=1`
`therefore" "e_(2)=sqrt5`
`"and "e_(1)=(2)/(sqrt5)`
`therefore" "e_(2):e_(1)=5:2`