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Find the equation of the hyperbola where...

Find the equation of the hyperbola where foci are `(0,+-12)`and the length of the latus rectum is 36.

Text Solution

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Here, Foci of hyperbola `= (0,+-12)`
That means the transverse axis of the hyperbola is `Y`-axis.
So, the equation will be of the type,
`y^2/a^2-x^2/b^2 = 1->(1)`
Also, `c = 12`
Length of latus rectum ` = 36`
`:. 2b^2/a = 36=> b^2 = 18a`
In a hyperbola, `c^2 = a^2+b^2`
...
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Knowledge Check

  • The equation of the hyperbola whose foci are (pm5,0) and length of the latus rectum is (9)/(2) is

    A
    `(x^(2))/(16) -(y^(2))/(9) =1`
    B
    `(x^(2))/(9) -(y^(2))/(16) =1`
    C
    `16x^(2) - 9y^(2) =1`
    D
    `9y^(2) - 16x^(2) =1`
  • The equation of hyperbola whose coordinates of the foci are (pm8,0) and the length of the latus rectum is 24 units. Is

    A
    `3x^(2)-y^(2)=48`
    B
    `4x^(2)-y^(2)=48`
    C
    `x^(2)-3y^(2)=48`
    D
    `x^(2)-4y^(2)=48`
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