Suppose an ellipse and a hyperbola have the same pair of foci on the `x`-axis with centres at the origin and they intersect at `(2,2)`. If the eccentricity of the ellipse is `(1)/(2)`, then the eccentricity of the hyperbola, is

Options

A. `sqrt((7)/(4))`
B. `sqrt((7)/(3))`
C. `sqrt((5)/(4))`
D. `sqrt((5)/(3))`

Explanation

Let the equations of the ellipse and hyperbola be `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` and `(x^(2))/(A^(2))-(y^(2))/(B^(2))=1` respectively. These two pass through `(2,2)`.
`(1)/(a^(2))+(1)/(b^(2))=(1)/(4)`…….`(i)`
and , `(1)/(A^(2))-(1)/(B^(2))=(1)/(4)`......`(ii)`
Let `e_(1)` and `e_(2)` be the eccentricities of the ellipse and hyperbola respectively. Then,
`b^(2)=a^(2)(1-e_(1)^(2))`.......`(iii)` and, `B^(2)=A^(2)(e_(2)^(2)-1)`.....`(iv)`
It is given that `e_(1)=(1)/(2)` and the ellipse and hyperbola have the same pair of foci on the `x`-axis.
`:.ae_(1)=Ae_(2)impliesAe_(2)=(a)/(2)`[`:'e_(1)=(1)/(2)`] .......`(v)`
Putting `e_(1)=(1)/(2)` in `(iii)`, we get
`b^(2)=(3a^(2))/(4)implies(1)/(b^(2))=(4)/(3a^(2))`........`(vi)`
From `(i)` and `(vi)`, we get
`(1)/(a^(2))+(4)/(3a^(2))=(1)/(4)implies(7)/(3a^(2))=(1)/(4)impliesa^(2)=(28)/(3)`
Putting `a^(2)=(28)/(3)` in `(vi)`, we get
`(1)/(b^(2))=(4)/(3)xx(3)/(28)impliesb^(2)=7`
Putting `a=2sqrt((7)/(3))` in `(v)`, we get
`sqrt((7)/(3))implies3A^(2)e_(2)^(2)=7`.......`(vii)`
From `(iv)` and `(vii)`, we get
`B^(2)=(7)/(3)-A^(2)`
Putting this value of `B^(2)` in `(ii)`, we get
`(1)/(A^(2))-(3)/(7-3A^(2))impliesA^(2)=1`
`:. B^(2)=(4)/(3)`
Hence, `e_(2)^(2)=1+(B^(2))/(A^(2))impliese_(2)^(2)=1+(4)/(3)=(7)/(3)impliese_(2)=sqrt((7)/(3))`