SOLUTION: Given the hyperbola: 9x2 – 54x – 16y2 + 64y – 127 = 0 a) Put the equation in standard form. b) Horizontal or vertical transverse axis?

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Question 975429: Given the hyperbola: 9x2 – 54x – 16y2 + 64y – 127 = 0

a) Put the equation in standard form.

b) Horizontal or vertical transverse axis?
c) Find center, vertices, foci, and the
equation of both asymptotes.
d) Sketch a graph of this hyperbola.

Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
Given the hyperbola: 9x2 – 54x – 16y2 + 64y – 127 = 0
a) Put the equation in standard form.
b) Horizontal or vertical transverse axis?
c) Find center, vertices, foci, and the
equation of both asymptotes.
d) Sketch a graph of this hyperbola.
***
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

This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation: , (h,k)=coordinates of center
center: (3, 2)
a^2=16
a=4
vertices: (3ħa, 2)=(3ħ4, 2)=(-1, 2) and (7, 2)
b^2=9
b=3
c^2=a^2+b^2=16+9=25
c=5
foci: (3ħc, 2)=(3ħ5, 2)=(-2, 2) and (8, 2)
Asymptotes: (2 lines that intersect and go thru center)
slopes of asymptotes=ħb/a=ħ3/4
..
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
..
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
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