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Find the equation of the hyperbola whose...

Find the equation of the hyperbola whose foci are (8, 3) and (0, 3) and eccentricity is `4/3` .

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Centre of hyper set`((8+0)/2,(3+3)/2)`
Let 2a and 2b be the length of transverse and conjugate axis of hyperbola.
Let e be the receipt for equation of hyper.
`(x-4)^2/a^2+(y-3)^2/b^2=1`
`S_1S_2=sqrt((b-a)^2+(c-a)^2`
`b^2=a^2(e^2-1)`
`b^2=9(16/9-1)`
`=16-9=7`
...
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Knowledge Check

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

    A
    `12(x -1)^(2) - 4(y-4)^(2) = 75`
    B
    `12(x+1)^(2) -4(y+4)^(2) = 75`
    C
    `4(x - 1)^(2) - 1( y- 4)^(2) = 75`
    D
    `4(x + 1)^(2) - 12(y + 4)^(2) = 75`

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

    A
    `(x+1)^(2)/(16)-(y-5)^(2)/(9)=1`
    B
    `x^(2)/(16)-y^(2)/(2)=1`
    C
    `(x-1)^(2)/(16)-(y-5)^(2)/(9)=-1`
    D
    none of these

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

    A
    `x ^(2) - (y-4) ^(2) =5 `
    B
    ` (x ^(2))/(9) - (4 (y-4)^(2))/(7) =1`
    C
    `(x ^(2))/(9) -(y ^(2))/(7) =1/4`
    D
    None of these
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