Question

The foci of hyperbola $9x^2-16y^2+18x+32y-151=0$, are

• A. $(2,3)$, $(5,7)$
• B. $(4,1)$, $(-6,1)$
• C. $(0,0),$(5,2)$
• D. none of these

Answer 1

Answer: (A)

$(2,3)$, $(5,7)$

Hyperbola:
9x^2-16y^2-18x-32y-151=0
9x^2-18x-16y^2-32y-151=0
complete the square: (you already did with a small error which I have corrected)
This is a hyperbola with horizontal transverse axis: (x-term listed first)
Its standard form of equation: , (h,k)=(x,y) coordinates of center.
Center:(1,-1)
a^2=16
a=√16=4
Vertices: (1a,-1)=(14,-1)=(-3,-1) and (5-1)
..
b^2=9
b=√9=3
..
c^2=a^2+b^2=16+9=25
c=√25=5
Foci: (1c,-1)=(15,-1)=(-4,-1) and (6-1)
..
Asymptotes:(straight line equations that go thru center, of the form y=mx+b, m=slope, b=y-intercept)
For hyperbolas with horizontal transverse axis: slope=b/a=3/4

Equation for asymptote with negative slope:
y=-3x/4+b
solve for b using coordinates of center.
-1=-3*1/4+b
b=-1/4
equation:y=-3x/4-1/4
..
Equation for asymptote with positive slope:
y=3x/4+b
solve for b using coordinates of center.
-1=3*1/4+b
b=-7/4

equation:y=3x/4-5/4
..
Graph:
9(x^2-2x+1)-16(y^2+2y+1)=151-16+9
9(x-1)^2-16(y+1)^2=144
(x-1)^2/16-(y+1)^2/9=1
..
See graph below:
y=((9(x-1)^2/16)-9)^.5-1