The foci of a hyperbola coincide with the foci of the ellipse `(x^2)/(25)+(y^2)/9=1.` Find the equation of the hyperbola, if its eccentricity is 2.

Options

A. `3x^(2)-y^(2)=6`
B. `3x^(2)-y^(2)=12`
C. `x^(2)-3y^(2)=12`
D. none of these

Answer

B

Explanation

The equation of the ellipse is of the form `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, where `a^(2)=25` and `b^(2)=9`.
Let `e` be the eccentricity of the ellipse. Then,
`e=sqrt(1-(b^(2))/(a^(2)))impliese=sqrt(1-(9)/(25))=(4)/(5)`
So, the coordinates of foci are `(+-ae,0)` i.e. `(+-4,0)`.
It is given that the foci of the hyperbola coincide with the foci of the ellipse. So, the coordinates of foci of the hyperbola are `(+-4,0)`.
Let `e'` be the eccentricity of the required hyperbola and its equation be
`(x^(2))/(a'^(2))-(y^(2))/(b'^(2))=1`.......`(i)`
The coordinates of foci are `(+-a'e',0)`
`:.a'e'=4implies2a'=4impliesa'=2` `[ :'e'=2`]
Also,
`b'^(2)=a'^(2)(e'^(2)-1)impliesb'^(2)=4(4-1)=12`
Subsituting the value of `a'` and `b'` in `(i)`, we get
`(x^(2))/(4)-(y^(2))/(12)=1` as the equation of the required hyperbola.