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Pre-calc

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Heres the equation:
(x^2)-(4y^2)-(4x)+(24y)-(36)=0
Your supposed to find the coordinates of the center, foci, and vertices, and the equations of the asymptotes of its graph.
So first you complete the square right?
(x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
[((x+2)^2)/76] -[(y+3)^2)/19]=1
So its a hyperbola, and is the center at (-2, -3)??

  • Pre-calc - ,

    Picking it up from your

    (x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
    should have been
    (x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
    notice on the left you had
    ... 4(..... +9) , so you subtracted 36, thus you must subtract 36 from the right side as well
    (you had added 36)

    final version

    (x-2)^2 - 4(y-3)^2 = 36

    or

    (x-2)^2 /36 - (y-3)^2 /9 = 1

    centre would be (2,3)
    a = 1 , b = 3 , c = √10

    vertices: (1,3) and (3,3)
    foci :( 2-√10 , 3) and (2+√10 , 3)
    asymptotes:

    first: y = 3x + b , with (2,3) on it
    3 = 6+b
    b = -3 ----> y = 3x - 3
    second: y = -3x + b
    3 = -6+b
    b = 9 -----> y = -3x + 9

  • Pre-calc - ,

    Watch your signs!
    (x^2 -4x+4)-4(y^2 +6y+9)=36+4+36
    should be (x^2 -4x+4)-4(y^2 -6y+9)=36+4-36
    (x^2 -4x+4)/4-4(y^2 -6y+9)/4=4/4
    ((x-2)^2)/4- (y-3)^2 =1
    hyperbola, center at (2,3)

  • correction -Pre-calc - ,

    from
    (x-2)^2 /36 - (y-3)^2 /9 = 1 , the rest should say

    centre is (2,3)
    a = 6, b = 3, c = √45


    vertices: (-4,3) and (8,3)
    foci :( 2-√45 , 3) and (2+√45 , 3)
    asymptotes:

    first: y = (1/2)x + b , with (2,3) on it
    3 = 1+b
    b = 2 ----> y = (1/2)x + 2
    second: y = -(1/2)x + b
    3 = 1+b
    b = 2 -----> y = -(1/2)x + 2

    check my arithmetic

  • Pre-calc - ,

    Thank you guys so much :)

  • Argggh - correction #2Pre-calc - ,

    I am sorry, don't know where I got that -4 on the RS from
    Go with Nadine's final equation
    (x^2 -4x+4)-4(y^2 -6y+9)=36+4-36

    ((x-2)^2)/4- (y-3)^2 =1

    from there, a=2, b = 1, c = √5

    adjust from there

  • Pre-calc - ,

    Haha ok i totally get it thanks!

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