Question 975429
Given the hyperbola: 9x2 – 54x – 16y2 + 64y – 127 = 0 
a) Put the equation in standard form. {{{(x-3)^2/16-(y-2)^2/9 =1}}}
b) Horizontal or vertical transverse axis? 
c) Find center, vertices, foci, and the 
equation of both asymptotes. 
d) Sketch a graph of this hyperbola.
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9x^2 – 54x – 16y^2 + 64y – 127 = 0 
complete the square:
9(x^2–6x+9)–16(y^2-4y+4) = 127+81-64
9(x-3)^2–16(y-2)^2 =144
{{{(x-3)^2/16-(y-2)^2/9 =1}}}
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=coordinates of center
center: (3, 2)
a^2=16
a=4
vertices: (3±a, 2)=(3±4, 2)=(-1, 2) and (7, 2)
b^2=9
b=3
c^2=a^2+b^2=16+9=25
c=5
foci: (3±c, 2)=(3±5, 2)=(-2, 2) and (8, 2)
Asymptotes: (2 lines that intersect  and go thru center)
slopes of asymptotes=±b/a=±3/4
..
equation of asymptote with negative slope:
y=-3x/4+b
solve for b using coordinates (3, 2) of center
2=-3*3/4+b
b=2+9/4=17/4
equation: y=-3x/4+17/4
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equation of asymptote with positive slope slope:
y=3x/4+b
solve for b using coordinates (3, 2) of center
2=3*3/4+b
b=2-9/4=-1/4
equation: y=3x/4-1/4

see graph below:
y=((9/16)(x-2)^2-1)^.5+2
{{{ graph( 300, 300, -10, 10, -10, 10,((9/16)(x-3)^2-1)^.5+2,-((9/16)(x-3)^2-1)^.5+2,-3x/4+17/4,3x/4-1/4) }}}