A hyperbola having the transverse axis of length `(1)/(2)` unit is confocal with the ellipse `3x^(2)+4y^(2) = 12`, then

Options

A. Equation of the hyperbola is `(x^(2))/(15)-(y^(2))/(1) = (1)/(16)`
B. Eccentricity of the hyperbola is 4
C. Distance between the directries of the hyperbola is `(1)/(8)` units
D. Length of latus rectum of the hyperbola is `(15)/(2)` units

Answer

B, C, D

Explanation

Ellipse is `(x^(2))/(4)+(y^(2))/(3) =1`
Here `(3)/(4) =1 - e^(2) rArr e = (1)/(2)`
Foci are `(+-1,0)`
Now the hyperbola is having the same focus, i.e., `(+-1,0)`.
Let `e_(1)` be the accentricity of hyperbola
`2ae_(1) =2`
But `2a = (1)/(2)` So, `e_(1) = 4`
`b^(2) = a^(2) (e_(1)^(2)-1) = (1)/(16) (16-1) = (15)/(16)`
So, the equation of the hyperbola is
`(x^(2))/((1)/(16)) - (y^(2))/((15)/(16)) = 1 rArr (x^(2))/(1) - (y^(2))/(15) = (1)/(16)`
Its distance between the directrixes `= (2a)/(e_(1)) = (1)/(2xx4) = (1)/(8)` units
Length of latus-rectum `= (2b^(2))/(a) = (2xx 15 xx 4)/(16 xx 1) = (15)/(2)` units