Lesson Hyperbola (concept and graphing calc.)
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Algebra: Conic sections - ellipse, parabola, hyperbola
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Hyperbolas, as you most likely already know, are two identical parabolas opening in opposite directions. There are two main ones known as: horizontal and vertical. The transverse axis is a straight line that connects the two parabolas. The distance of the transverse axis is defined as {{{2a}}}. The conjugate axis is the axis perpendicular to the transverse axis and goes through the center. This axis's distance is defined as {{{2b}}}. This makes an imaginery box which can be used to better help those who draw hyperbolas on a grid by freelance. Standard Format: {{{1 = (x - h)^2/a^2 - (y - k)^2/b^2}}} ~> For a horizontal transverse axis. Standard Format: {{{1 = (y - k)^2/a^2 - (x - h)^2/b^2}}} ~> For a vertical transverse axis. Center: (h,k) First, we will work with the standard form for a horizontal transverse axis. {{{1 = (x - h)^2/a^2 - (y - k)^2/b^2}}} {{{1 - (x - h)^2/a^2 = -(y - k)^2/b^2}}} {{{-1 + (x - h)^2/a^2 = (y - k)^2/b^2}}} {{{b^2(-1 + (x - h)^2/a^2) = (y - k)^2}}} +-{{{sqrt(b^2(-1 + (x - h)^2/a^2)) = sqrt((y - k)^2)}}} +-{{{b*sqrt((-1 + (x - h)^2/a^2)) = y - k}}} +-{{{b*sqrt((-1 + (x - h)^2/a^2)) + k = y}}} +-{{{b*sqrt((-a^2/a^2 + (x - h)^2/a^2)) + k = y}}} +-{{{b*sqrt((-a^2 + (x - h)^2)/a^2) + k = y}}} +-{{{(b/a)*sqrt(-a^2 + (x - h)^2) + k = y}}} Ironically, +-b/a is the slope for the asymptotes for this hyperbola. Asymptote for horizontal hyperbolas: y - y1 = m(x - x1) for the center (h,k) y - k = (+-b/a)(x - h) which is: y - k = (b/a)(x - h) and y - k = (-b/a)(x - h) y - k = bx/a - bh/a and y - k = -bx/a + hb/a y = bx/a - bh/a + k and y = -bx/a + hb/a + k Now, lets see how this works: Center: (2,1) Transverse Axis: 4 horizontally ~> 2a = 4 or a = 2 Conjugate Axis: 6 vertically ~> 2b = 6 or b = 3 +-{{{(b/a)*sqrt(-a^2 + (x - h)^2) + k = y}}} +-{{{(3/2)*sqrt(-4 + (x - 2)^2) + 1 = y}}} Asymptote: y = bx/a - bh/a + k and y = -bx/a + hb/a + k y = 3x/2 - 3(2)/2 + 1 and y = -3x/2 + (2)3/2 + 1 y = 3x/2 - 3 + 1 and y = -3x/2 + 3 + 1 y = 3x/2 - 2 and y = -3x/2 + 4 Now, the graphing: {{{graph(300,300,-10,10,-10,10,-(3/2)*sqrt(-4 + (x - 2)^2) + 1,(3/2)*sqrt(-4 + (x - 2)^2) + 1,-3x/2 + 4,3x/2 - 2)}}} It looks successful.... Now, we will work with the standard form for a vertical transverse axis. {{{1 = (y - k)^2/a^2 - (x - h)^2/b^2}}} {{{1 + (x - h)^2/b^2 = (y - k)^2/a^2}}} {{{a^2(1 + (x - h)^2/b^2) = (y - k)^2}}} {{{sqrt(a^2(1 + (x - h)^2/b^2)) = sqrt((y - k)^2)}}} +-{{{a*sqrt((1 + (x - h)^2/b^2)) = y - k}}} +-{{{a*sqrt((1 + (x - h)^2/b^2)) + k = y}}} +-{{{a*sqrt((b^2/b^2 + (x - h)^2/b^2)) + k = y}}} +-{{{a*sqrt((b^2 + (x - h)^2)/b^2) + k = y}}} +-{{{(a/b)*sqrt(b^2 + (x - h)^2) + k = y}}} Ironically again, +-a/b is the slope for the asymptotes for this hyperbola. Asymptote for vertical hyperbolas: y - y1 = m(x - x1) for center point (h,k) y - k = (+-a/b)(x - h) y - k = (a/b)(x - h) or y - k = (-a/b)(x - h) y - k = ax/b - ah/b or y - k = -ax/b + ah/b y = ax/b - ah/b + k or y = -ax/b + ah/b + k Now, lets see how this works: Center: (-2,-3) Transverse Axis: 8 vertically ~> 2a = 8 or a = 4 Conjugate Axis: 6 horizontally ~> 2b = 6 or b = 3 +-{{{(a/b)*sqrt(b^2 + (x - h)^2) + k = y}}} +-{{{(4/3)*sqrt(9 + (x + 2)^2) - 3 = y}}} Asymptote: y = ax/b - ah/b + k or y = -ax/b + ah/b + k y = 4x/3 - 4(-2)/3 - 3 or y = -4x/3 + 4(-2)/3 - 3 y = 4x/3 + 8/3 - 9/3 or y = -4x/3 - 8/3 - 9/3 y = 4x/3 - 1/3 or y = -4x/3 - 17/3 Now, the graphing: {{{graph(300,300,-10,10,-10,10,(4/3)*sqrt(9 + (x + 2)^2) - 3,(-4/3)*sqrt(9 + (x + 2)^2) - 3,-4x/3 - 17/3,4x/3 - 1/3)}}}