Consider an ellipse ` x^2/a^2+y^2/b^2=1` Let a hyperbola is having its vertices at the extremities of minor axis of an ellipse and length of major axis of an ellipse is equal to the distance between the foci of hyperbola. Let `e_1` and `e_2` be the eccentricities of an ellipse and hyperbola respectively. Again let A be the area of the quadrilateral formed by joining all the foci and A, be the area of the quadrilateral formed by all the directrices. The relation between `e_1 and e_2` is given by

Options

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

Answer

B

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`