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The focus and the correspondin directrix...

The focus and the correspondin directrix of a hyperbola are (1-3) and y=2 and eccentricity is 3/2. Find the equation of the hyperbola.

Text Solution

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The correct Answer is:
`4x^(2)-5y^(2)-8x+60y+4=0`
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Knowledge Check

  • One of the foci of the hyperbola is origin and the corresponding directrix is 3x + 4y + 1=0 .The eccentricity of the hyperbola is sqrt 5 .The equation of the hyperbola is

    A
    ` 4x ^(2) + 11y ^(2) + 24xy + 6x + 8y + 1=0`
    B
    ` 8x^(2) + 9y^(2) + 24 xy + 6x + 6y+1 =0`
    C
    ` 8x^(2) + 9y^(2) + 24xy + 6x+ 8y +1=0 `
    D
    ` 8x^(2) + 9y^(2) -24xt + 6x + 8y + 1=0`

  • i: The equation of the conic with focus at (1,-1) directrix along x-y+1 =0 and with eccentricity sqrt2 is 2xy -4x+ 4y + 1=0 II : One of the foci of the hyperbola is origin and the corresponding directrix is 3x+ 4y + 1=0 .The eccentricity of the hyperbola sqrt 5 .The equation of the hyperbola is 4x^(2) +11y ^(2) + 24xy + 6x+ 8y + 1=0

    A
    only I is true
    B
    only II is true
    C
    both I and II are ture
    D
    neither I nor II true

  • The equation of the hyperbola whose foci are ( +- 5,0) and eccentricity 5/3 is

    A
    ` (x^(2))/( 9) -y^( 2)/( 16) =1`
    B
    ` ( x^(2) )/( 4) -( y^(2))/( 16) = 1`
    C
    ` (x^(2) )/( 9) +( y^(2)) /( 6) = 1 `
    D
    ` (x^(2))/( 4) +(y^(2))/( 16) = 1 `

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    One focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is the line y = 2. Find the equation of the hyperbola if its eccentricity is (3)/(2)

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    If the eccentricity of a hyperbola is (5)/(3) , then the eccentricity of its conjugate hyperbola is

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