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Find foci, vertices and asymptotes of the hyperbola.

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the asymptotes, vertices, and foci of the hyperbola. 4x^2-y^2-24x-4y+28=0

    2. Relevant equations
    (x-h)^2/a^2-(y-k)^2/b^2=1, asymptotes= k± (b/a)(x-h), vertices ± a from center, foci ± c from center, c^2=a^2+b^2, center= (h,k).


    3. The attempt at a solution

    I begin by rewriting the equation of 4x^2-y^2-24x-4y+28=0.
    as: 4x^2-24x-y^2-4y=-28
    Then I complete the square for each segment (x/y) which ends up as 4(x-3)^2-9 and -(y+2)^2-4.
    Rewrite this into the orginal equation as 4(x-3)^2-(y+2)^2=-28+36-4
    4(x-3)^2-(y+2)^2=4
    I then rewrite this in the form of (x-h)^2/a^2-(y-k)^2/b^2=1
    thus (x-3)^2/1-(y+2)^2/4=1
    I have now found that a^2=1 and b^2=4 thus, a=1, and b=2
    Now I solve for the questions asked
    I have the center at (h,k) being (3, -2)
    C^2=a^2+b^2, =1+4=5, C=√5
    foci are equal to (3±√ 5, -2)
    a=1 vertices (4,-2) & (2,-2)
    lastly the asymptotes for the form k ± b/a (x-h)
    would be in the form of -2 ± 2/1 (x-3), -2 ± 2(x-3)

    That is all my work/solutions for this problem.
    Any help with what I may have done wrong and how to learn how to correct is, or even just saying "correct" is appreciated thank you.
     
  2. jcsd
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