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The equation of the hyperbol whose foci ...

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

A

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

B

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

C

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

D

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

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • The equation to the ellipse whose foci are (pm2,0) and eccentricity (1)/(2) is :

    A
    `(x^(2))/(12)+(y^(2))/(16)=1`
    B
    `(x^(2))/(16)+(y^(2))/(12)=1`
    C
    `(x^(2))/(16)+(y^(2))/(8)=1`
    D
    None of these

  • The equation of the ellipse whose one focus is at (4,0) and whose eccentricity is (4)/(5) is

    A
    ` (x^(2))/(3^(2))+(y^(2))/(5^(2))=1`
    B
    `(x^(2))/(5^(2))+(y^(2))/(3^(2))=1`
    C
    `(x^(2))/(5^(2))+(y^(2))/(4^(2))=1`
    D
    ` (x^(2))/(4^(2))+(y^(2))/(5^(2))=1`

  • The equation of the hyperbola with vertices at (pm5,0) and foci at (pm7,0) is :

    A
    `-x^2/25+y^2/24=1`
    B
    `x^2/25-y^2/24=1`
    C
    `x^2/24-y^2/25=1`
    D
    `-x^2/24+y^2/25=1`

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    The equation of the hyperbola with eccentricity 3/2 and foci at (pm2,0) is

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    The equation of the ellipse whose vertices are (-4,1) and (6,1) and a focus lies on x-2 y-2 =0 is

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