Lesson Hyperbola (concept and graphing calc.)

Source code of 'Hyperbola (concept and graphing calc.)'

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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)}}}