If the ellipse `x^(2)+2y^(2)=4` and the hyperbola S = 0 have same end points of the latus rectum, then the eccentricity of the hyperbola can be

Options

A. `"cosec"(pi)/(4)`
B. `"cosec"(pi)/(3)`
C. `2sin.(pi)/(3)+sin.(pi)/(4)`
D. `sqrt2 sin.(pi)/(3)+sin.(pi)/(4)`

Answer

A, D

Explanation

(1), (4)
End point of latus rectum of ellipse are `(pm sqrt2, pm1)`.
Let hyperbola be `(x^(2))/(A^(2))-(y^(2))/(B^(2))=1`. ltBrgt End points of latus rectum of hyperbola coincide with those of ellipse.
`therefore" "Ae=sqrt2 and (B^(2))/(A)=1`
Also, `B^(2)=A^(2)(e^(2)-1)`
`rArr" "A=A^(2)e^(2)-A^(2)`
Eliminating B, we get
`rArr" "A^(2)+A-2=0`
`therefore A = 1 and e = sqrt2`
Let hyperbola be `(y^(2))/(B^(2))-(x^(2))/(A^(2))=1`.
`therefore" "Be=1, (A^(2))/(B)=sqrt2`
Also, `e=sqrt(1+(A^(2))/(B^(2))`
`therefore" "(1)/(B^(2))=1+(sqrt2)/(B)`
`therefore" "B^(2)+sqrt2B-1=0`
`therefore" "B=(-sqrt2+sqrt6)/(2)`
`therefore" "e=(2)/(sqrt2(sqrt3-1))=(sqrt2(sqrt3+1))/(2)=(sqrt3+1)/(sqrt2)`