The equation of the hyperbola whose foci...

The equation of the hyperbola whose foci are (-2, 0) and (2,0) and eccentricity is 2 is given by

`3x^(2)-y^(2)=3`

`-x^(2)+3y^(2)=3`

`-3x^(2)+y^(2)=3`

`x^(2)-3y^(2)=3`

Text Solution

Verified by Experts

Let the equatin of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`,
where `b^(2)=a^(2)(e^(2)-1)`.
It is given that `ae=2` and `e=2` . Therefore , `a=1`.
Consequently, `b^(2)=1(4-1)=3`.
Hence, the equation of the hyperbola is
`(x^(2))/(1)-(y^(2))/(3)=1`, or `3x^(2)-y^(2)=3`.

Topper's Solved these Questions

CARTESIAN CO-ORDINATE SYSTEM
OBJECTIVE RD SHARMA|Exercise Exercise|29 Videos
DETERMINANTS
OBJECTIVE RD SHARMA|Exercise Chapter Test|30 Videos

The equation of the hyperbola, whose foci are (6, 4) and (-4, 4) and eccentricity is 2, is

`12(x -1)^(2) - 4(y-4)^(2) = 75`

`12(x+1)^(2) -4(y+4)^(2) = 75`

`4(x - 1)^(2) - 1( y- 4)^(2) = 75`

`4(x + 1)^(2) - 12(y + 4)^(2) = 75`

The equation of hyperbola whose foci are (2,4) and (-2,4) and eccentricity is 4/3, is

`x ^(2) - (y-4) ^(2) =5 `

` (x ^(2))/(9) - (4 (y-4)^(2))/(7) =1`

`(x ^(2))/(9) -(y ^(2))/(7) =1/4`

None of these

The equation of the hyperbola whose foci are (6,5),(-4,5) and eccentricity 5//4 is

`((x-1)^(2))/(16)-((y-5)^(2))/(9)=1`

`(x^(2))/(16)-(y^(2))/(9)=1`

`((x-1)^(2))/(16)-((y-5)^(2))/(9)=-1`

none of these