SOLUTION: Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci. y^2/16 - x^2/9 = 1

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Question 616682: Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci.
y^2/16 - x^2/9 = 1

Found 2 solutions by lwsshak3, ewatrrr:
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
Sketch the hyperbolas. Identify the center, asymptotes, vertices, and foci.
y^2/16 - x^2/9 = 1
**
Equation is that of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of the center
For given equation:
center: (0,0)
a^2=16
a=√16=4
vertices: (0, 0±a)=(0,0±4)=(0,-4) and (0,4)
..
b^2=9
b=√9=3
..
c^2=a^2+b^2=16+9=25
c=√25=5
foci:(0, 0±c)=(0,0±5)=(0,-5) and (0,5)
...
slopes of asymptotes with vertical transverse axis=±a/b=±4/3
asymptotes are straight lines which intersect at the center.
equation:y=mx+b, m=slope, b=y-intercept
y-intercept=0, since asymptotes go thru the origin(0,0)
equation of asymptote with negative slope: y=-4x/3
equation of asymptote with positive slope: y=4x/3
..
see graph below:
y=±(16+16x^2/9)^.5


Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
or
center, (0,0)
asymptotes, y = ± 4/3x
vertices, (0,-4), (0,4)
and foc(0,-5),(0,5) ||Note: = 5


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